# Ch. - 4 Principle of Mathematical Induction

## Chapter 4 Ex.4.1 Question 1

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

\begin{align}1 + 3 + {3^2} + \ldots + {3^{n - 1}} = \frac{{\left( {{3^n} - 1} \right)}}{2}.\end{align}

### Solution

Let $$P\left( n \right)$$ be the given statement.

i.e., \begin{align}P\left( n \right):1 + 3 + {3^2} + \ldots + {3^{n - 1}} = \frac{{\left( {{3^n} - 1} \right)}}{2}\end{align}

For $$n = 1$$,

\begin{align} P\left( 1 \right):1 = \frac{{\left( {{3^1} - 1} \right)}}{2} = \frac{2}{2} = 1\end{align}, which is true.

Assume that $$P\left( k \right)$$ is true for some positive integer $$k$$

i.e., \begin{align}1 + 3 + {3^2} + \ldots + {3^{k - 1}} = \frac{{\left( {{3^k} - 1} \right)}}{2} \qquad \ldots \left( 1 \right)\end{align}

We will now prove that $$P\left( {k + 1} \right)$$ is also true.

Now, we have

\begin{align}&1 + 3 + {3^2} + \ldots + {3^{\left( {k + 1} \right) - 1}}\\\Rightarrow \;&\left( {1 + 3 + {3^2} + \ldots + {3^{k - 1}}} \right) + {3^k}\\\Rightarrow \;&\frac{{\left( {{3^k} - 1} \right)}}{2} + {3^k} \qquad \qquad \ldots \left[ {{\rm{from}}\;\left( {\rm{1}} \right)} \right]\\\Rightarrow \;&\frac{{{3^k} - 1 + 2 \times {3^k}}}{2}\\\Rightarrow \;&\frac{{\left( {1 + 2} \right){3^k} - 1}}{2}\\\Rightarrow\; &\frac{{3 \times {3^k} - 1}}{2}\\\Rightarrow \;&\frac{{{3^{k + 1}} - 1}}{2}\end{align}

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 2

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

${1^3} + {2^3} + {3^3} + \ldots + {n^3} = {\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2}.$

### Solution

Let $$P\left( n \right)$$ be the given statement.

i.e., \begin{align}P\left( n \right):{1^3} + {2^3} + {3^3} + \ldots + {n^3} = {\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2} \end{align}

For $$n = 1$$,

\begin{align}P\left( 1 \right):{1^3} = 1 = {\left[ {\frac{{1\left( {1 + 1} \right)}}{2}} \right]^2} = {\left[ {\frac{{1 \times 2}}{2}} \right]^2} = {\left[ 1 \right]^2} = 1 \end{align}, which is true.

Assume that $$P\left( k \right)$$ is true for some positive integer $$k$$

i.e., \begin{align}{1^3} + {2^3} + {3^3} + \ldots + {k^3} = {\left[ {\frac{{k\left( {k + 1} \right)}}{2}} \right]^2} \qquad \quad \ldots \left( 1 \right) \end{align}

We will now prove that $$P\left( {k + 1} \right)$$ is also true.

Now, we have

\begin{align}&{1^3} + {2^3} + {3^3} + \ldots + {\left( {k + 1} \right)^3}\\\Rightarrow \;& \left( {{1^3} + {2^3} + {3^3} + \ldots + {k^3}} \right) + {\left( {k + 1} \right)^3}\\\Rightarrow \;&{\left[ {\frac{{k\left( {k + 1} \right)}}{2}} \right]^2} + {\left( {k + 1} \right)^3} \qquad \quad \ldots \left[ {{\rm{from}}\;\left( {\rm{1}} \right)} \right]\\\Rightarrow \;&\frac{{{k^2}{{\left( {k + 1} \right)}^2}}}{4} + {\left( {k + 1} \right)^3}\\\Rightarrow \;&\frac{{{k^2}{{\left( {k + 1} \right)}^2} + 4{{\left( {k + 1} \right)}^3}}}{4}\\\Rightarrow \;&\frac{{{{\left( {k + 1} \right)}^2}\left[ {{k^2} + 4\left( {k + 1} \right)} \right]}}{4}\\\Rightarrow \;&\frac{{{{\left( {k + 1} \right)}^2}\left[ {{k^2} + 4k + 4} \right]}}{4}\\\Rightarrow \;&\frac{{{{\left( {k + 1} \right)}^2}{{\left( {k + 2} \right)}^2}}}{4}\\\Rightarrow \;&{\left[ {\frac{{\left( {k + 1} \right)\left( {k + 2} \right)}}{2}} \right]^2}\\\Rightarrow \;&{\left[ {\frac{{\left( {k + 1} \right)\left( {k + 1 + 1} \right)}}{2}} \right]^2}\end{align}

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 3

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

\begin{align}1 + \frac{1}{{\left( {1 + 2} \right)}} + \frac{1}{{\left( {1 + 2 + 3} \right)}} + \ldots + \frac{1}{{\left( {1 + 2 + 3 + \ldots n} \right)}} = \frac{{2n}}{{\left( {n + 1} \right)}}.\end{align}

### Solution

Let $$P\left( n \right)$$ be the given statement.

i.e., \begin{align}P\left( n \right):1 + \frac{1}{{\left( {1 + 2} \right)}} + \frac{1}{{\left( {1 + 2 + 3} \right)}} + \ldots + \frac{1}{{\left( {1 + 2 + 3 + \ldots n} \right)}} = \frac{{2n}}{{\left( {n + 1} \right)}} \end{align}

For $$n = 1$$,

\begin{align}P\left( 1 \right):1 = \frac{{2 \times 1}}{{\left( {1 + 1} \right)}} = \frac{2}{2} = 1\end{align}, which is true.

Assume that $$P\left( k \right)$$ is true for some positive integer $$k$$

i.e., \begin{align}1 + \frac{1}{{\left( {1 + 2} \right)}} + \frac{1}{{\left( {1 + 2 + 3} \right)}} + \ldots + \frac{1}{{\left( {1 + 2 + 3 + \ldots k} \right)}} = \frac{{2k}}{{\left( {k + 1} \right)}} \qquad \quad \ldots \left( 1 \right) \end{align}

We will now prove that $$P\left( {k + 1} \right)$$ is also true.

Now, we have

\begin{align}& \ \ \ \ 1+\frac{1}{\left( 1+2 \right)}+\frac{1}{\left( 1+2+3 \right)}+\ldots +\frac{1}{\left( 1+2+3+\ldots \left( k+1 \right) \right)} \\& \Rightarrow \left[ 1+\frac{1}{\left( 1+2 \right)}+\frac{1}{\left( 1+2+3 \right)}+\ldots +\frac{1}{\left( 1+2+3+\ldots k \right)} \right]\\& \quad+\frac{1}{\left( 1+2+3+\ldots +k+\left( k+1 \right) \right)} \\& \Rightarrow \frac{2k}{\left( k+1 \right)}+\frac{1}{\left( 1+2+3+\ldots +k+\left( k+1 \right) \right)} \qquad \qquad \qquad\ldots \left[ \text{from}\ \left( \text{1} \right) \right] \\& \Rightarrow \frac{2k}{\left( k+1 \right)}+\frac{1}{\frac{\left( k+1 \right)\left( k+2 \right)}{2}} \quad \ldots \quad \left[ \because 1+2+3+\ldots +n=\frac{n\left( n+1 \right)}{2} \right] \\& \Rightarrow \frac{2k}{\left( k+1 \right)}+\frac{2}{\left( k+1 \right)\left( k+2 \right)} \\& \Rightarrow \frac{2k\left( k+2 \right)+2}{\left( k+1 \right)\left( k+2 \right)} \\& \Rightarrow \frac{2\left( {{k}^{2}}+2k+1 \right)}{\left( k+1 \right)\left( k+2 \right)} \\& \Rightarrow \frac{2{{\left( k+1 \right)}^{2}}}{\left( k+1 \right)\left( k+2 \right)} \\& \Rightarrow \frac{2\left( k+1 \right)}{\left( k+2 \right)} \\& \Rightarrow \frac{2\left( k+1 \right)}{\left( k+1 \right)+1}\end{align}

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 4

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

\begin{align}1.2.3 + 2.3.4 + \ldots + n\left( {n + 1} \right)\left( {n + 2} \right) = \frac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{4}.\end{align}

### Solution

Let $$P\left( n \right)$$ be the given statement.

i.e., \begin{align}P\left( n \right):1.2.3 + 2.3.4 + \ldots + n\left( {n + 1} \right)\left( {n + 2} \right) = \frac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{4} \end{align}

For $$n = 1$$,

\begin{align}P\left( 1 \right):1.2.3 = 6 = \frac{{1\left( {1 + 1} \right)\left( {1 + 2} \right)\left( {1 + 3} \right)}}{4} = \frac{{1.2.3.4}}{4} = 6 \end{align}, which is true.

Assume that $$P\left( k \right)$$ is true for some positive integer $$k$$ i.e.,

$1.2.3 + 2.3.4 + \ldots + k\left( {k + 1} \right)\left( {k + 2} \right) = \frac{{k\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}{4} \qquad \quad \ldots \left( 1 \right)$

We will now prove that $$P\left( {k + 1} \right)$$ is also true.

Now, we have

\begin{align}&1.2.3 + 2.3.4 + \ldots + \left( {k + 1} \right)\left[ {\left( {k + 1} \right) + 1} \right]\left[ {\left( {k + 1} \right) + 2} \right]\\\Rightarrow\; &\left[ {1.2.3 + 2.3.4 + \ldots + k\left( {k + 1} \right)\left( {k + 2} \right)} \right] + \left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)\\\Rightarrow \;&\frac{{k\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}{4} + \left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right) \qquad \qquad \ldots \left[ {{\rm{from}}\;\left( {\rm{1}} \right)} \right]\\\Rightarrow \;&\frac{{k\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right) + 4\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}{4}\\\Rightarrow \;&\frac{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)\left( {k + 4} \right)}}{4}\\\Rightarrow \;&\frac{{\left( {k + 1} \right)\left[ {\left( {k + 1} \right) + 1} \right]\left[ {\left( {k + 1} \right) + 2} \right]\left[ {\left( {k + 1} \right) + 3} \right]}}{4}\end{align}

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 5

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

$1.3 + {2.3^2} + {3.3^3} + \ldots + n{.3^n} = \frac{{\left( {2n - 1} \right){3^{n + 1}} + 3}}{4}.$

### Solution

Let $$P\left( n \right)$$ be the given statement.

i.e., \begin{align}P\left( n \right):1.3 + {2.3^2} + {3.3^3} + \ldots + n{.3^n} = \frac{{\left( {2n - 1} \right){3^{n + 1}} + 3}}{4} \end{align}

For $$n = 1$$,

\begin{align}P\left( 1 \right):1.3 = 3 = \frac{{\left( {2 \times 1 - 1} \right){3^{1 + 1}} + 3}}{4} = \frac{{{{1.3}^2} + 3}}{4} = \frac{{12}}{4} = 3\end{align}, which is true.

Assume that $$P\left( k \right)$$ is true for some positive integer k

i.e., \begin{align}1.3 + {2.3^2} + {3.3^3} + \ldots + k{.3^k} = \frac{{\left( {2k - 1} \right){3^{k + 1}} + 3}}{4} \qquad \quad \ldots \left( 1 \right)\end{align}

We will now prove that $$P\left( {k + 1} \right)$$ is also true.

Now, we have

\begin{align}&1.3 + {2.3^2} + {3.3^3} + \ldots + \left( {k + 1} \right){.3^{k + 1}}\\\Rightarrow \;&\left[ {1.3 + {{2.3}^2} + {{3.3}^3} + \ldots + k{{.3}^k}} \right] + \left( {k + 1} \right){.3^{k + 1}}\\\Rightarrow \;&\frac{{\left( {2k - 1} \right){3^{k + 1}} + 3}}{4} + \left( {k + 1} \right){.3^{k + 1}} \qquad \quad \ldots \left[ {{\rm{from}}\;\left( {\rm{1}} \right)} \right]\\\Rightarrow \;&\frac{{\left( {2k - 1} \right){3^{k + 1}} + 3 + 4\left( {k + 1} \right){{.3}^{k + 1}}}}{4}\\\Rightarrow \;&\frac{{{3^{k + 1}}\left[ {\left( {2k - 1} \right) + 4\left( {k + 1} \right)} \right] + 3}}{4}\\\Rightarrow \;&\frac{{{3^{k + 1}}\left[ {2k - 1 + 4k + 4} \right] + 3}}{4}\\\Rightarrow \;&\frac{{{3^{k + 1}}\left[ {6k + 3} \right] + 3}}{4}\\\Rightarrow \;&\frac{{{3^{k + 1}}.3\left[ {2k + 1} \right] + 3}}{4}\\\Rightarrow \;&\frac{{\left[ {2k + 2 - 1} \right]{{.3}^{\left( {k + 1} \right) + 1}} + 3}}{4}\\\Rightarrow \;&\frac{{\left[ {2\left( {k + 1} \right) - 1} \right]{3^{\left( {k + 1} \right) + 1}} + 3}}{4}\end{align}

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 6

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

$1.2 + 2.3 + 3.4 + \ldots + n.\left( {n + 1} \right) = \left[ {\frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{3}} \right].$

### Solution

Let $$P\left( n \right)$$ be the given statement.

i.e., \begin{align}P\left( n \right):1.2 + 2.3 + 3.4 + \ldots + n.\left( {n + 1} \right) = \left[ {\frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{3}} \right] \end{align}

For $$n = 1$$,

\begin{align}P\left( 1 \right):1.2 = 2 = \left[ {\frac{{1\left( {1 + 1} \right)\left( {1 + 2} \right)}}{3}} \right] = \frac{{1.2.3}}{3} = 2 \end{align}, which is true.

Assume that $$P\left( k \right)$$ is true for some positive integer $$k$$ i.e.,

\begin{align}1.2 + 2.3 + 3.4 + \ldots + k.\left( {k + 1} \right) = \left[ {\frac{{k\left( {k + 1} \right)\left( {k + 2} \right)}}{3}} \right] \qquad\qquad \ldots \left( 1 \right) \end{align}

We will now prove that $$P\left( {k + 1} \right)$$ is also true.

Now, we have

\begin{align}&1.2 + 2.3 + 3.4 + \ldots + \left( {k + 1} \right)\left[ {\left( {k + 1} \right) + 1} \right]\\\Rightarrow \;&\left[ {1.2 + 2.3 + 3.4 + \ldots + k.\left( {k + 1} \right)} \right] + \left( {k + 1} \right)\left( {k + 2} \right)\\\Rightarrow \;&\left[ {\frac{{k\left( {k + 1} \right)\left( {k + 2} \right)}}{3}} \right] + \left( {k + 1} \right)\left( {k + 2} \right) \qquad \quad\ldots \left[ {{\rm{from}}\;\left( {\rm{1}} \right)} \right]\\\Rightarrow \;&\frac{{k\left( {k + 1} \right)\left( {k + 2} \right) + 3\left( {k + 1} \right)\left( {k + 2} \right)}}{3}\\\Rightarrow \;&\frac{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}{3}\\\Rightarrow \;&\frac{{\left( {k + 1} \right)\left[ {\left( {k + 1} \right) + 1} \right]\left[ {\left( {k + 1} \right) + 2} \right]}}{3}\end{align}

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 7

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

\begin{align}1.3 + 3.5 + 5.7 + \ldots + \left( {2n - 1} \right)\left( {2n + 1} \right) = \frac{{n\left( {4{n^2} + 6n - 1} \right)}}{3}. \end{align}

### Solution

Let $$P\left( n \right)$$ be the given statement.

i.e., \begin{align}1.3 + 3.5 + 5.7 + \ldots + \left( {2n - 1} \right)\left( {2n + 1} \right) = \frac{{n\left( {4{n^2} + 6n - 1} \right)}}{3}. \end{align}

For $$n = 1$$,

\begin{align}P\left( 1 \right):1.3 = 3 = \frac{{1\left( {{{4.1}^2} + 6.1 - 1} \right)}}{3} = \frac{9}{3} = 3\end{align}, which is true.

Assume that $$P\left( k \right)$$ is true for some positive integer $$k$$ i.e.,

\begin{align}1.3 + 3.5 + 5.7 + \ldots + \left( {2k - 1} \right)\left( {2k + 1} \right) = \frac{{k\left( {4{k^2} + 6k - 1} \right)}}{3} \qquad \quad \ldots \left( 1 \right) \end{align}

We will now prove that $$P\left( {k + 1} \right)$$ is also true.

Now, we have

\begin{align}&1.3 + 3.5 + 5.7 + \ldots + \left[ {2\left( {k + 1} \right) - 1} \right]\left[ {2\left( {k + 1} \right) + 1} \right]\\\Rightarrow \;&\left[ {1.3 + 3.5 + 5.7 + \ldots + \left( {2k - 1} \right)\left( {2k + 1} \right)} \right] + \left( {2k + 1} \right)\left( {2k + 3} \right)\\\Rightarrow \;&\left[ {\frac{{k\left( {4{k^2} + 6k - 1} \right)}}{3}} \right] + \left( {4{k^2} + 8k + 3} \right) \qquad \qquad \ldots \left[ {{\rm{from}}\;\left( {\rm{1}} \right)} \right]\\\Rightarrow \;&\frac{{k\left( {4{k^2} + 6k - 1} \right) + 3\left( {4{k^2} + 8k + 3} \right)}}{3}\\\Rightarrow \;&\frac{{4{k^3} + 6{k^2} - k + 12{k^2} + 24k + 9}}{3}\\\Rightarrow \;&\frac{{4{k^3} + 18{k^2} + 23k + 9}}{3}\\\Rightarrow \;&\frac{{4{k^3} + 14{k^2} + 9k + 4{k^2} + 14k + 9}}{3}\\\Rightarrow \;&\frac{{k\left( {4{k^2} + 14k + 9} \right) + \left( {4{k^2} + 14k + 9} \right)}}{3}\\\Rightarrow \;&\frac{{\left( {k + 1} \right)\left( {4{k^2} + 14k + 9} \right)}}{3}\\\Rightarrow \;&\frac{{\left( {k + 1} \right)\left( {4{k^2} + 8k + 4 + 6k + 6 - 1} \right)}}{3}\\\Rightarrow \;&\frac{{\left( {k + 1} \right)\left[ {4\left( {{k^2} + 2k + 2} \right) + 6\left( {k + 1} \right) - 1} \right]}}{3}\\\Rightarrow \;&\frac{{\left( {k + 1} \right)\left[ {4{{\left( {k + 1} \right)}^2} + 6\left( {k + 1} \right) - 1} \right]}}{3}\end{align}

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 8

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

$1.2 + {2.2^2} + {3.2^3} + \ldots + n{.2^n} = \left( {n - 1} \right){2^{n + 1}} + 2.$

### Solution

Let $$P\left( n \right)$$ be the given statement.

i.e., \begin{align}P\left( n \right):1.2 + {2.2^2} + {3.2^3} + \ldots + n{.2^n} = \left( {n - 1} \right){2^{n + 1}} + 2\end{align}

For $$n = 1$$,

\begin{align}P\left( 1 \right):1.2 = 2 = \left( {1 - 1} \right){2^{1 + 1}} + 2 = 0 + 2 = 2\end{align}, which is true.

Assume that $$P\left( k \right)$$ is true for some positive integer $$k$$ i.e.,

$1.2 + {2.2^2} + {3.2^3} + \ldots + k{.2^n} = \left( {k - 1} \right){2^{k + 1}} + 2 \qquad \quad \ldots \left( 1 \right)$

We will now prove that $$P\left( {k + 1} \right)$$ is also true.

Now, we have

\begin{align}&1.2 + {2.2^2} + {3.2^3} + \ldots + \left( {k + 1} \right){.2^{k + 1}}\\\Rightarrow \;&\left[ {1.2 + {{2.2}^2} + {{3.2}^3} + \ldots + k{{.2}^n}} \right] + \left( {k + 1} \right){.2^{k + 1}}\\\Rightarrow\;& \left( {k - 1} \right){2^{k + 1}} + 2 + \left( {k + 1} \right){.2^{k + 1}} \qquad \quad \ldots \left[ {{\rm{from}}\;\left( {\rm{1}} \right)} \right]\\\Rightarrow\;& \left[ {\left( {k - 1} \right) + \left( {k + 1} \right)} \right]{2^{k + 1}} + 2\\\Rightarrow\;& 2k{.2^{\left( {k + 1} \right)}} + 2\\\Rightarrow \;&k{.2^{\left( {k + 1} \right) + 1}} + 2\\\Rightarrow\;& \left[ {\left( {k + 1} \right) - 1} \right]{.2^{\left( {k + 1} \right) + 1}} + 2\end{align}

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 9

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

\begin{align}\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{{{2^n}}} = 1 - \frac{1}{{{2^n}}}.\end{align}

### Solution

Let $$P\left( n \right)$$ be the given statement.

i.e., \begin{align}P\left( n \right):\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{{{2^n}}} = 1 - \frac{1}{{{2^n}}} \end{align}

For $$n = 1$$,

\begin{align}P\left( 1 \right):\frac{1}{2} = 1 - \frac{1}{{{2^1}}} = 1 - \frac{1}{2} = \frac{1}{2} \end{align}, which is true.

Assume that $$P\left( k \right)$$ is true for some positive integer $$k$$ i.e.,

\begin{align}\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{{{2^k}}} = 1 - \frac{1}{{{2^k}}} \qquad \quad \ldots \left( 1 \right) \end{align}

We will now prove that $$P\left( {k + 1} \right)$$ is also true.

Now, we have

\begin{align}&\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{{{2^{k + 1}}}}\\\Rightarrow\;& \left[ {\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{{{2^k}}}} \right] + \frac{1}{{{2^{k + 1}}}}\\\Rightarrow \;&1 - \frac{1}{{{2^k}}} + \frac{1}{{{2^{k + 1}}}} \qquad \qquad\ldots \left[ {{\rm{from}}\;\left( {\rm{1}} \right)} \right]\\\Rightarrow \;&1 - \frac{1}{{{2^k}}}\left( {1 - \frac{1}{2}} \right)\\\Rightarrow\;& 1 - \frac{1}{{{2^k}}}.\frac{1}{2}\\\Rightarrow \;&1 - \frac{1}{{{2^{k + 1}}}}\end{align}

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 10

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

\begin{align}\frac{1}{{2.5}} + \frac{1}{{5.8}} + \frac{1}{{8.11}} + \ldots + \frac{1}{{\left( {3n - 1} \right)\left( {3n + 2} \right)}} = \frac{n}{{\left( {6n + 4} \right)}}.\end{align}

### Solution

Let $$P\left( n \right)$$ be the given statement.

i.e., \begin{align}P\left( n \right):\frac{1}{{2.5}} + \frac{1}{{5.8}} + \frac{1}{{8.11}} + \ldots + \frac{1}{{\left( {3n - 1} \right)\left( {3n + 2} \right)}} = \frac{n}{{\left( {6n + 4} \right)}}\end{align}

For $$n = 1$$,

\begin{align}P\left( 1 \right):\frac{1}{{2.5}} = \frac{1}{{10}} = \frac{1}{{\left( {6.1 + 4} \right)}} = \frac{1}{{10}}\end{align}, which is true.

Assume that $$P\left( k \right)$$ is true for some positive integer $$k$$ i.e.,

\begin{align}\frac{1}{{2.5}} + \frac{1}{{5.8}} + \frac{1}{{8.11}} + \ldots + \frac{1}{{\left( {3k - 1} \right)\left( {3k + 2} \right)}} = \frac{k}{{\left( {6k + 4} \right)}} \qquad \quad \ldots \left( 1 \right)\end{align}

We will now prove that $$P\left( {k + 1} \right)$$ is also true.

Now, we have

\begin{align}&\frac{1}{{2.5}} + \frac{1}{{5.8}} + \frac{1}{{8.11}} + \ldots + \frac{1}{{\left[ {3\left( {k + 1} \right) - 1} \right]\left[ {3\left( {k + 1} \right) + 2} \right]}}\\\Rightarrow \;& \left[ {\frac{1}{{2.5}} + \frac{1}{{5.8}} + \frac{1}{{8.11}} + \ldots + \frac{1}{{\left( {3k - 1} \right)\left( {3k + 2} \right)}}} \right] + \frac{1}{{\left( {3k + 2} \right)\left( {3k + 5} \right)}}\\\Rightarrow \;&\frac{k}{{\left( {6k + 4} \right)}} + \frac{1}{{\left( {3k + 2} \right)\left( {3k + 5} \right)}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left[ {{\rm{from}}\;\left( {\rm{1}} \right)} \right]\\\Rightarrow \;&\frac{k}{{2\left( {3k + 2} \right)}} + \frac{1}{{\left( {3k + 2} \right)\left( {3k + 5} \right)}}\\\Rightarrow \;&\frac{1}{{\left( {3k + 2} \right)}}\left[ {\frac{k}{2} + \frac{1}{{\left( {3k + 5} \right)}}} \right]\\\Rightarrow \;&\frac{1}{{\left( {3k + 2} \right)}}\left[ {\frac{{k\left( {3k + 5} \right) + 2}}{{2\left( {3k + 5} \right)}}} \right]\\\Rightarrow \;&\frac{1}{{\left( {3k + 2} \right)}}\left[ {\frac{{3{k^2} + 5k + 2}}{{2\left( {3k + 5} \right)}}} \right]\\\Rightarrow \;&\frac{1}{{\left( {3k + 2} \right)}}\left[ {\frac{{3{k^2} + 3k + 2k + 2}}{{2\left( {3k + 5} \right)}}} \right]\\\Rightarrow \;&\frac{1}{{\left( {3k + 2} \right)}}\left[ {\frac{{3k\left( {k + 1} \right) + 2\left( {k + 1} \right)}}{{2\left( {3k + 5} \right)}}} \right]\\\Rightarrow \;&\frac{1}{{\left( {3k + 2} \right)}}\left[ {\frac{{\left( {k + 1} \right)\left( {3k + 2} \right)}}{{2\left( {3k + 5} \right)}}} \right]\\\Rightarrow \;&\frac{{\left( {k + 1} \right)}}{{\left( {6k + 10} \right)}}\\\Rightarrow \;&\frac{{\left( {k + 1} \right)}}{{\left[ {\left( {6k + 6} \right) + 4} \right]}}\\\Rightarrow \;&\frac{{\left( {k + 1} \right)}}{{\left[ {6\left( {k + 1} \right) + 4} \right]}}\end{align}

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 11

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

\begin{align}\frac{1}{{1.2.3}} + \frac{1}{{2.3.4}} + \frac{1}{{3.4.5}} + \ldots + \frac{1}{{n\left( {n + 1} \right)\left( {n + 2} \right)}} = \frac{{n\left( {n + 3} \right)}}{{4\left( {n + 1} \right)\left( {n + 2} \right)}}. \end{align}

### Solution

Let $$P\left( n \right)$$ be the given statement.

i.e., \begin{align}P\left( n \right):\frac{1}{{1.2.3}} + \frac{1}{{2.3.4}} + \frac{1}{{3.4.5}} + \ldots + \frac{1}{{n\left( {n + 1} \right)\left( {n + 2} \right)}} = \frac{{n\left( {n + 3} \right)}}{{4\left( {n + 1} \right)\left( {n + 2} \right)}}\end{align}

For $$n = 1$$,

\begin{align}P\left( 1 \right):\frac{1}{{1.2.3}} = \frac{1}{6} = \frac{{1\left( {1 + 3} \right)}}{{4\left( {1 + 1} \right)\left( {1 + 2} \right)}} = \frac{{1.4}}{{4.2.3}} = \frac{1}{6}\end{align}, which is true.

Assume that $$P\left( k \right)$$ is true for some positive integer $$k$$ i.e.,

\begin{align}\frac{1}{{1.2.3}} + \frac{1}{{2.3.4}} + \frac{1}{{3.4.5}} + \ldots + \frac{1}{{k\left( {k + 1} \right)\left( {k + 2} \right)}} = \frac{{k\left( {k + 3} \right)}}{{4\left( {k + 1} \right)\left( {k + 2} \right)}} \qquad \qquad \ldots \left( 1 \right)\end{align}

We will now prove that $$P\left( {k + 1} \right)$$ is also true.

Now, we have

\begin{align}&\frac{1}{{1.2.3}} + \frac{1}{{2.3.4}} + \frac{1}{{3.4.5}} + \ldots + \frac{1}{{\left( {k + 1} \right)\left[ {\left( {k + 1} \right) + 1} \right]\left[ {\left( {k + 1} \right) + 2} \right]}}\\\Rightarrow \;&\left[ {\frac{1}{{1.2.3}} + \frac{1}{{2.3.4}} + \frac{1}{{3.4.5}} + \ldots + \frac{1}{{k\left( {k + 1} \right)\left( {k + 2} \right)}}} \right]\\& + \frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}\\\Rightarrow \;&\frac{{k\left( {k + 3} \right)}}{{4\left( {k + 1} \right)\left( {k + 2} \right)}} + \frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}} \qquad \qquad\ldots \left[ {{\rm{from}}\;\left( {\rm{1}} \right)} \right]\\\Rightarrow \;&\frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\left[ {\frac{{k\left( {k + 3} \right)}}{4} + \frac{1}{{\left( {k + 3} \right)}}} \right]\\\Rightarrow\;& \frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\left[ {\frac{{k{{\left( {k + 3} \right)}^2} + 4}}{{4\left( {k + 3} \right)}}} \right]\\\Rightarrow \;&\frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\left[ {\frac{{k\left( {{k^2} + 6k + 9} \right) + 4}}{{4\left( {k + 3} \right)}}} \right]\\\Rightarrow\;& \frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\left[ {\frac{{{k^3} + 6{k^2} + 9k + 4}}{{4\left( {k + 3} \right)}}} \right]\\\Rightarrow\;& \frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\left[ {\frac{{{k^3} + 2{k^2} + k + 4{k^2} + 8k + 4}}{{4\left( {k + 3} \right)}}} \right]\\\Rightarrow\;& \frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\left[ {\frac{{k\left( {{k^2} + 2k + 1} \right) + 4\left( {{k^2} + 2k + 1} \right)}}{{4\left( {k + 3} \right)}}} \right]\\\Rightarrow \;&\frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\left[ {\frac{{\left( {k + 4} \right)\left( {{k^2} + 2k + 1} \right)}}{{4\left( {k + 3} \right)}}} \right]\\\Rightarrow \;&\frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\left[ {\frac{{\left( {k + 4} \right){{\left( {k + 1} \right)}^2}}}{{4\left( {k + 3} \right)}}} \right]\\\Rightarrow \;&\frac{{\left( {k + 1} \right)\left( {k + 4} \right)}}{{4\left( {k + 2} \right)\left( {k + 3} \right)}}\\\Rightarrow\;& \frac{{\left( {k + 1} \right)\left[ {\left( {k + 1} \right) + 3} \right]}}{{4\left[ {\left( {k + 1} \right) + 1} \right]\left[ {\left( {k + 1} \right) + 2} \right]}}\end{align}

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 12

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

\begin{align}a + ar + a{r^2} + \ldots + a{r^{n - 1}} = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}.\end{align}

### Solution

Let $$P\left( n \right)$$ be the given statement.

i.e., \begin{align}P\left( n \right):a + ar + a{r^2} + \ldots + a{r^{n - 1}} = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\end{align}

For $$n = 1$$,

\begin{align}P\left( 1 \right):a = \frac{{a\left( {{r^1} - 1} \right)}}{{r - 1}} = \frac{{a\left( {r - 1} \right)}}{{r - 1}} = a\end{align}, which is true.

Assume that $$P\left( k \right)$$ is true for some positive integer $$K$$ i.e.,

\begin{align}a + ar + a{r^2} + \ldots + a{r^{k - 1}} = \frac{{a\left( {{r^k} - 1} \right)}}{{r - 1}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

We will now prove that $$P\left( {k + 1} \right)$$ is also true.

Now, we have\begin{align} &a + ar + a{r^2} + \ldots + a{r^{\left( {k + 1} \right) - 1}}\\\Rightarrow \;& \left[ {a + ar + a{r^2} + \ldots + a{r^{k - 1}}} \right] + a{r^k}\\\Rightarrow \;&\frac{{a\left( {{r^k} - 1} \right)}}{{r - 1}} + a{r^k} \qquad \quad \ldots \left[ {{\rm{from}}\;\left( {\rm{1}} \right)} \right]\\\Rightarrow \;&\frac{{a\left( {{r^k} - 1} \right) + a{r^k}\left( {r - 1} \right)}}{{r - 1}}\\\Rightarrow \;&\frac{{a{r^k} - a + a{r^{k + 1}} - a{r^k}}}{{r - 1}}\\\Rightarrow \;&\frac{{a{r^{k + 1}} - a}}{{r - 1}}\\\Rightarrow \;&\frac{{a\left( {{r^{k + 1}} - 1} \right)}}{{r - 1}}\end{align}

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 13

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

\begin{align}\left( {1 + \frac{3}{1}} \right)\left( {1 + \frac{5}{4}} \right)\left( {1 + \frac{7}{9}} \right) \ldots \left( {1 + \frac{{\left( {2n + 1} \right)}}{{{n^2}}}} \right) = {\left( {n + 1} \right)^2}.\end{align}

### Solution

Let $$P\left( n \right)$$ be the given statement.

i.e., \begin{align}P\left( n \right):\left( {1 + \frac{3}{1}} \right)\left( {1 + \frac{5}{4}} \right)\left( {1 + \frac{7}{9}} \right) \ldots \left( {1 + \frac{{\left( {2n + 1} \right)}}{{{n^2}}}} \right) = {\left( {n + 1} \right)^2}\end{align}

For $$n = 1$$,

\begin{align}P\left( 1 \right):\left( {1 + \frac{3}{1}} \right) = 4 = {\left( {1 + 1} \right)^2} = {2^2} = 4\end{align}, which is true.

Assume that $$P\left( k \right)$$ is true for some positive integer $$k$$ i.e.,

\begin{align}\left( {1 + \frac{3}{1}} \right)\left( {1 + \frac{5}{4}} \right)\left( {1 + \frac{7}{9}} \right) \ldots \left( {1 + \frac{{\left( {2k + 1} \right)}}{{{k^2}}}} \right) = {\left( {k + 1} \right)^2} \qquad \quad \ldots \left( 1 \right) \end{align}

We will now prove that $$P\left( {k + 1} \right)$$ is also true.

Now, we have

\begin{align}&\left( {1 + \frac{3}{1}} \right)\left( {1 + \frac{5}{4}} \right)\left( {1 + \frac{7}{9}} \right) \ldots \left( {1 + \frac{{\left[ {2\left( {k + 1} \right) + 1} \right]}}{{{{\left( {k + 1} \right)}^2}}}} \right)\\\Rightarrow \;&\left[ {\left( {1 + \frac{3}{1}} \right)\left( {1 + \frac{5}{4}} \right)\left( {1 + \frac{7}{9}} \right) \ldots \left( {1 + \frac{{\left( {2k + 1} \right)}}{{{k^2}}}} \right)} \right]\left( {1 + \frac{{\left( {2k + 3} \right)}}{{{{\left( {k + 1} \right)}^2}}}} \right)\\\Rightarrow\;& {\left( {k + 1} \right)^2}\left( {1 + \frac{{\left( {2k + 3} \right)}}{{{{\left( {k + 1} \right)}^2}}}} \right) \qquad \quad \ldots \left[ {{\rm{from}}\;\left( {\rm{1}} \right)} \right]\\\Rightarrow \;&{\left( {k + 1} \right)^2}\left( {\frac{{{{\left( {k + 1} \right)}^2} + \left( {2k + 3} \right)}}{{{{\left( {k + 1} \right)}^2}}}} \right)\\\Rightarrow\;& {\left( {k + 1} \right)^2} + \left( {2k + 3} \right)\\\Rightarrow \;&{k^2} + 2k + 1 + 2k + 3\\\Rightarrow \;& {k^2} + 4k + 4\\\Rightarrow\;& {\left( {k + 2} \right)^2}\\\Rightarrow\;& {\left( {k + 1 + 1} \right)^2}\end{align}

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 14

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

\begin{align}\left( {1 + \frac{1}{1}} \right)\left( {1 + \frac{1}{2}} \right)\left( {1 + \frac{1}{3}} \right) \ldots \left( {1 + \frac{1}{n}} \right) = \left( {n + 1} \right). \end{align}

### Solution

Let $$P\left( n \right)$$ be the given statement.

i.e., \begin{align}P\left( n \right):\left( {1 + \frac{1}{1}} \right)\left( {1 + \frac{1}{2}} \right)\left( {1 + \frac{1}{3}} \right) \ldots \left( {1 + \frac{1}{n}} \right) = \left( {n + 1} \right)\end{align}

For $$n = 1$$,

\begin{align}P\left( 1 \right):\left( {1 + \frac{1}{1}} \right) = 2 = \left( {1 + 1} \right) = 2\end{align}, which is true.

Assume that $$P\left( k \right)$$ is true for some positive integer $$k$$ i.e.,

\begin{align}\left( {1 + \frac{1}{1}} \right)\left( {1 + \frac{1}{2}} \right)\left( {1 + \frac{1}{3}} \right) \ldots \left( {1 + \frac{1}{k}} \right) = \left( {k + 1} \right) \qquad \quad \ldots \left( 1 \right) \end{align}

We will now prove that $$P\left( {k + 1} \right)$$ is also true.

Now, we have

\begin{align}&\left( {1 + \frac{1}{1}} \right)\left( {1 + \frac{1}{2}} \right)\left( {1 + \frac{1}{3}} \right) \ldots \left( {1 + \frac{1}{{k + 1}}} \right)\\\Rightarrow\;& \left[ {\left( {1 + \frac{1}{1}} \right)\left( {1 + \frac{1}{2}} \right)\left( {1 + \frac{1}{3}} \right) \ldots \left( {1 + \frac{1}{k}} \right)} \right]\left( {1 + \frac{1}{{k + 1}}} \right)\\\Rightarrow \;&\left( {k + 1} \right)\left( {1 + \frac{1}{{k + 1}}} \right)\\\Rightarrow \;& \left( {k + 1} \right)\left( {\frac{{k + 1 + 1}}{{k + 1}}} \right) \qquad \quad \ldots \left[ {{\rm{from}}\;\left( {\rm{1}} \right)} \right]\\\Rightarrow \;& \left[ {\left( {k + 1} \right) + 1} \right]\end{align}

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 15

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

\begin{align}{1^2} + {3^2} + {5^2} + \ldots + {\left( {2n - 1} \right)^2} = \frac{{n\left( {2n - 1} \right)\left( {2n + 1} \right)}}{3}. \end{align}

### Solution

Let $$P\left( n \right)$$ be the given statement.

i.e., \begin{align}P\left( n \right):{1^2} + {3^2} + {5^2} + \ldots + {\left( {2n - 1} \right)^2} = \frac{{n\left( {2n - 1} \right)\left( {2n + 1} \right)}}{3} \end{align}

For $$n = 1$$,

\begin{align}P\left( 1 \right):{1^2} = 1 = \frac{{1\left( {2.1 - 1} \right)\left( {2.1 + 1} \right)}}{3} = \frac{{1.1.3}}{3} = 1 \end{align}, which is true.

Assume that $$P\left( k \right)$$ is true for some positive integer $$k$$ i.e.,

\begin{align}{1^2} + {3^2} + {5^2} + \ldots + {\left( {2k - 1} \right)^2} = \frac{{k\left( {2k - 1} \right)\left( {2k + 1} \right)}}{3} \qquad \quad\ldots \left( 1 \right) \end{align}

We will now prove that $$P\left( {k + 1} \right)$$ is also true.

Now, we have

\begin{align}&{1^2} + {3^2} + {5^2} + \ldots + {\left[ {2\left( {k + 1} \right) - 1} \right]^2}\\\Rightarrow\;& \left[ {{1^2} + {3^2} + {5^2} + \ldots + {{\left( {2k - 1} \right)}^2}} \right] + {\left( {2k + 1} \right)^2}\\\Rightarrow\;& \frac{{k\left( {2k - 1} \right)\left( {2k + 1} \right)}}{3} + {\left( {2k + 1} \right)^2} \qquad \quad \ldots \left[ {{\rm{from}}\;\left( {\rm{1}} \right)} \right]\\\Rightarrow \;& \frac{{k\left( {2k - 1} \right)\left( {2k + 1} \right) + 3{{\left( {2k + 1} \right)}^2}}}{3}\\\Rightarrow \;&\frac{{\left( {2k + 1} \right)\left[ {k\left( {2k - 1} \right) + 3\left( {2k + 1} \right)} \right]}}{3}\\\Rightarrow \;&\frac{{\left( {2k + 1} \right)\left[ {2{k^2} - k + 6k + 3} \right]}}{3}\\\Rightarrow \;&\frac{{\left( {2k + 1} \right)\left[ {2{k^2} + 5k + 3} \right]}}{3}\\\Rightarrow \;& \frac{{\left( {2k + 1} \right)\left[ {2{k^2} + 2k + 3k + 3} \right]}}{3}\\\Rightarrow\;& \frac{{\left( {2k + 1} \right)\left[ {2k\left( {k + 1} \right) + 3\left( {k + 1} \right)} \right]}}{3}\\\Rightarrow \;&\frac{{\left( {2k + 1} \right)\left( {k + 1} \right)\left( {2k + 3} \right)}}{3}\\\Rightarrow \;&\frac{{\left( {k + 1} \right)\left[ {2\left( {k + 1} \right) - 1} \right]\left[ {2\left( {k + 1} \right) + 1} \right]}}{3}\end{align}

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 16

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

\begin{align}\frac{1}{{1.4}} + \frac{1}{{4.7}} + \frac{1}{{7.10}} + \ldots + \frac{1}{{\left( {3n - 2} \right)\left( {3n + 1} \right)}} = \frac{n}{{\left( {3n + 1} \right)}}.\end{align}

### Solution

Let $$P\left( n \right)$$ be the given statement.

i.e., \begin{align}P\left( n \right):\frac{1}{{1.4}} + \frac{1}{{4.7}} + \frac{1}{{7.10}} + \ldots + \frac{1}{{\left( {3n - 2} \right)\left( {3n + 1} \right)}} = \frac{n}{{\left( {3n + 1} \right)}}\end{align}

For $$n = 1$$,

\begin{align}P\left( 1 \right):\frac{1}{{1.4}} = \frac{1}{4} = \frac{1}{{\left( {3.1 + 1} \right)}} = \frac{1}{4}\end{align}, which is true.

Assume that $$P\left( k \right)$$ is true for some positive integer $$k$$ i.e.,

\begin{align}\frac{1}{{1.4}} + \frac{1}{{4.7}} + \frac{1}{{7.10}} + \ldots + \frac{1}{{\left( {3k - 2} \right)\left( {3k + 1} \right)}} = \frac{k}{{\left( {3k + 1} \right)}} \qquad \quad \ldots \left( 1 \right) \end{align}

We will now prove that $$P\left( {k + 1} \right)$$ is also true.

Now, we have

\begin{align}&\frac{1}{{1.4}} + \frac{1}{{4.7}} + \frac{1}{{7.10}} + \ldots + \frac{1}{{\left[ {3\left( {k + 1} \right) - 2} \right]\left[ {3\left( {k + 1} \right) + 1} \right]}}\\\Rightarrow \;&\left[ {\frac{1}{{1.4}} + \frac{1}{{4.7}} + \frac{1}{{7.10}} + \ldots + \frac{1}{{\left( {3k - 2} \right)\left( {3k + 1} \right)}}} \right]\\& + \frac{1}{{\left( {3k + 1} \right)\left( {3k + 4} \right)}}\\\Rightarrow \;&\frac{k}{{\left( {3k + 1} \right)}} + \frac{1}{{\left( {3k + 1} \right)\left( {3k + 4} \right)}} \qquad \qquad \ldots \left[ {{\rm{from}}\;\left( {\rm{1}} \right)} \right]\\\Rightarrow \;&\frac{{k\left( {3k + 4} \right) + 1}}{{\left( {3k + 1} \right)\left( {3k + 4} \right)}}\\\Rightarrow \;&\frac{{3{k^2} + 4k + 1}}{{\left( {3k + 1} \right)\left( {3k + 4} \right)}}\\\Rightarrow \;&\frac{{3{k^2} + 3k + k + 1}}{{\left( {3k + 1} \right)\left( {3k + 4} \right)}}\\\Rightarrow \;&\frac{{3k\left( {k + 1} \right) + \left( {k + 1} \right)}}{{\left( {3k + 1} \right)\left( {3k + 4} \right)}}\\\Rightarrow\;& \frac{{\left( {3k + 1} \right)\left( {k + 1} \right)}}{{\left( {3k + 1} \right)\left( {3k + 4} \right)}}\\\Rightarrow \;&\frac{{\left( {k + 1} \right)}}{{\left[ {3\left( {k + 1} \right) + 1} \right]}}\end{align}

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 17

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

\begin{align}\frac{1}{{3.5}} + \frac{1}{{5.7}} + \frac{1}{{7.9}} + \ldots + \frac{1}{{\left( {2n + 1} \right)\left( {2n + 3} \right)}} = \frac{n}{{3\left( {2n + 3} \right)}}.\end{align}

### Solution

Let $$P\left( n \right)$$ be the given statement.

i.e., \begin{align}P\left( n \right):\frac{1}{{3.5}} + \frac{1}{{5.7}} + \frac{1}{{7.9}} + \ldots + \frac{1}{{\left( {2n + 1} \right)\left( {2n + 3} \right)}} = \frac{n}{{3\left( {2n + 3} \right)}}\end{align}

For $$n = 1$$,

\begin{align}P\left( 1 \right):\frac{1}{{3.5}} = \frac{1}{{15}} = \frac{1}{{3\left( {2.1 + 3} \right)}} = \frac{1}{{3.5}} = \frac{1}{{15}}\end{align}, which is true.

Assume that $$P\left( k \right)$$ is true for some positive integer $$k$$ i.e.,

\begin{align}\frac{1}{{3.5}} + \frac{1}{{5.7}} + \frac{1}{{7.9}} + \ldots + \frac{1}{{\left( {2k + 1} \right)\left( {2k + 3} \right)}} = \frac{k}{{3\left( {2k + 3} \right)}} \qquad \quad \ldots \left( 1 \right)\end{align}

We will now prove that $$P\left( {k + 1} \right)$$ is also true.

Now, we have

\begin{align}&\frac{1}{{3.5}} + \frac{1}{{5.7}} + \frac{1}{{7.9}} + \ldots + \frac{1}{{\left[ {2\left( {k + 1} \right) + 1} \right]\left[ {2\left( {k + 1} \right) + 3} \right]}}\\\Rightarrow \;&\left[ {\frac{1}{{3.5}} + \frac{1}{{5.7}} + \frac{1}{{7.9}} + \ldots + \frac{1}{{\left( {2k + 1} \right)\left( {2k + 3} \right)}}} \right] \\&+ \frac{1}{{\left( {2k + 3} \right)\left( {2k + 5} \right)}}\\\Rightarrow \;&\frac{k}{{3\left( {2k + 3} \right)}} + \frac{1}{{\left( {2k + 3} \right)\left( {2k + 5} \right)}} \qquad \qquad \ldots \left[ {{\rm{from}}\;\left( {\rm{1}} \right)} \right]\\\Rightarrow \;&\frac{{k\left( {2k + 5} \right) + 3}}{{3\left( {2k + 3} \right)\left( {2k + 5} \right)}}\\\Rightarrow \;&\frac{{2{k^2} + 5k + 3}}{{3\left( {2k + 3} \right)\left( {2k + 5} \right)}}\\\Rightarrow \;&\frac{{2{k^2} + 2k + 3k + 3}}{{3\left( {2k + 3} \right)\left( {2k + 5} \right)}}\\\Rightarrow \;& \frac{{2k\left( {k + 1} \right) + 3\left( {k + 1} \right)}}{{3\left( {2k + 3} \right)\left( {2k + 5} \right)}}\\\Rightarrow \;& \frac{{\left( {2k + 3} \right)\left( {k + 1} \right)}}{{3\left( {2k + 3} \right)\left( {2k + 5} \right)}}\\\Rightarrow \;& \frac{{\left( {k + 1} \right)}}{{3\left( {2k + 5} \right)}}\\\Rightarrow \;& \frac{{\left( {k + 1} \right)}}{{3\left[ {2\left( {k + 1} \right) + 3} \right]}}\end{align}

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 18

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

\begin{align}1 + 2 + 3 + \ldots + n < \frac{1}{8}{\left( {2n + 1} \right)^2}. \end{align}

### Solution

Let $$P\left( n \right)$$ be the given statement.

i.e., \begin{align}P\left( n \right):1 + 2 + 3 + \ldots + n < \frac{1}{8}{\left( {2n + 1} \right)^2}\end{align}

We note that $$P\left( n \right)$$ is true for $$n = 1$$,

Since,

\begin{align}P\left( 1 \right):1 < \frac{1}{8}{\left( {2.1 + 1} \right)^2} = \frac{9}{8} = 1\frac{1}{8}\end{align}

Assume that $$P\left( k \right)$$ is true for some positive integer $$k$$ i.e.,

\begin{align}1 + 2 + 3 + \ldots + k < \frac{1}{8}{\left( {2k + 1} \right)^2} \qquad \quad \ldots \left( 1 \right)\end{align}

We will now prove that $$P\left( {k + 1} \right)$$ is true whenever $$P\left( k \right)$$is true

Now, we have

\begin{align}1 + 2 + 3 + \ldots + k &< \frac{1}{8}{\left( {2k + 1} \right)^2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{from}}\;\left( {\rm{1}} \right)} \right]\\1 + 2 + 3 + \ldots + k + \left( {k + 1} \right)& < \frac{1}{8}{\left( {2k + 1} \right)^2} + \left( {k + 1} \right)\\&< \frac{1}{8}\left[ {{{\left( {2k + 1} \right)}^2} + 8\left( {k + 1} \right)} \right]\\&< \frac{1}{8}\left[ {4{k^2} + 4k + 1 + 8k + 8} \right]\\&< \frac{1}{8}\left[ {4{k^2} + 12k + 9} \right]\\&< \frac{1}{8}{\left[ {2k + 3} \right]^2}\\&< \frac{1}{8}{\left[ {2\left( {k + 1} \right) + 1} \right]^2}\end{align}

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 19

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

$$n\left( {n + 1} \right)\left( {n + 5} \right)$$ is a multiple of $$3.$$

### Solution

We can write

$$P\left( n \right):n\left( {n + 1} \right)\left( {n + 5} \right)$$ is a multiple of $$3.$$

We note that

$$P\left( 1 \right):1\left( {1 + 1} \right)\left( {1 + 5} \right) = 1.2.6 = 12$$ which is a multiple of $$3.$$

Thus $$P\left( n \right)$$ is true for $$n = 1$$

Let $$P\left( k \right)$$ be true for some natural number $$k,$$

i.e., $$P\left( k \right):k\left( {k + 1} \right)\left( {k + 5} \right)$$ is a multiple of $$3.$$

We can write

$k\left( {k + 1} \right)\left( {k + 5} \right) = 3a \qquad \quad \ldots \left( 1 \right)$

where $$a \in N$$ .

Now, we will prove that $$P\left( {k + 1} \right)$$ is true whenever $$P\left( k \right)$$ is true.

Now,

\begin{align}&\left( {k + 1} \right)\left[ {\left( {k + 1} \right) + 1} \right]\left[ {\left( {k + 1} \right) + 5} \right]\\\Rightarrow \;&\left( {k + 1} \right)\left( {k + 2} \right)\left[ {\left( {k + 5} \right) + 1} \right]\\\Rightarrow \;&\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 5} \right) + \left( {k + 1} \right)\left( {k + 2} \right)\\\Rightarrow \;&\left( {k + 2} \right)\left[ {\left( {k + 1} \right)\left( {k + 5} \right)} \right] + \left( {k + 1} \right)\left( {k + 2} \right)\\\Rightarrow \;&\left[ {k\left( {k + 1} \right)\left( {k + 5} \right) + 2\left( {k + 1} \right)\left( {k + 5} \right)} \right] + \left( {k + 1} \right)\left( {k + 2} \right)\\\Rightarrow \;& \left[ {3a + 2\left( {k + 1} \right)\left( {k + 5} \right)} \right] + \left( {k + 1} \right)\left( {k + 2} \right) \qquad \qquad \qquad \quad \left[ {{\rm{from}}\left( {\rm{1}} \right)} \right]\\\Rightarrow \;&3a + \left( {k + 1} \right)\left[ {2\left( {k + 5} \right) + \left( {k + 2} \right)} \right]\\\Rightarrow\;& 3a + \left( {k + 1} \right)\left[ {2k + 10 + k + 2} \right]\\\Rightarrow \;&3a + \left( {k + 1} \right)\left[ {3k + 12} \right]\\\Rightarrow \;&3a + 3\left( {k + 1} \right)\left( {k + 4} \right)\\\Rightarrow\;&3\left[ {a + \left( {k + 1} \right)\left( {k + 4} \right)} \right]\end{align}

From the last line, we see that

$$3\left[ {a + \left( {k + 1} \right)\left( {k + 4} \right)} \right]$$ is a multiple of $$3.$$

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 20

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

$${10^{2n - 1}} + 1$$ is divisible by $$11.$$

### Solution

We can write

$$P\left( n \right):{10^{2n - 1}} + 1$$ is divisible by $$11.$$

We note that

$$P\left( 1 \right):{10^{2.1 - 1}} + 1 = 10 + 1 = 11$$ which is divisible by $$11.$$

Thus $$P\left( n \right)$$ is true for $$n = 1$$

Let $$P\left( k \right)$$ be true for some natural number$$k,$$

i.e., $$P\left( k \right):{10^{2k - 1}} + 1$$ is divisible by $$11.$$

We can write

${10^{2k - 1}} + 1 = 11a \qquad \quad \ldots \left( 1 \right)$

where $$a \in N$$ .

Now, we will prove that $$P\left( {k + 1} \right)$$ is true whenever $$P\left( k \right)$$ is true.

Now,

\begin{align}&{10^{2\left( {k + 1} \right) - 1}} + 1\\\Rightarrow &\;{10^{2k + 1}} + 1\\\Rightarrow &\;{10^2}\left( {{{10}^{2k - 1}}} \right) + 1\\\Rightarrow &\;{10^2}\left( {{{10}^{2k - 1}} + 1 - 1} \right) + 1\\\Rightarrow &\;{10^2}\left( {{{10}^{2k - 1}} + 1} \right) - {10^2} + 1\\\Rightarrow &\;{10^2}.11a - 100 + 1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{from}}\left( {\rm{1}} \right)} \right]\\\Rightarrow &\;{10^2}.11a - 99\\\Rightarrow &\;11\left( {100a - 9} \right)\end{align}

From the last line, we see that

$$11\left( {100a - 9} \right)$$ is divisible by $$11.$$

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 21

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

$${x^{2n}} - {y^{2n}}$$ is divisible by $$x + y$$.

### Solution

We can write

$$P\left( n \right):{x^{2n}} - {y^{2n}}$$is divisible by $$x + y$$.

We note that

$$P\left( 1 \right):{x^{2.1}} - {y^{2.1}} = {x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right)$$ which is divisible by $$x + y$$.

Thus $$P\left( n \right)$$ is true for $$n = 1$$

Let $$P\left( k \right)$$ be true for some natural number $$k,$$

i.e., $$P\left( k \right):{x^{2k}} - {y^{2k}}$$ is divisible by $$x + y$$.

We can write

${x^{2k}} - {y^{2k}} = a\left( {x + y} \right) \qquad \qquad \ldots \left( 1 \right)$

where $$a \in N$$ .

Now, we will prove that $$P\left( {k + 1} \right)$$ is true whenever $$P\left( k \right)$$ is true.

Now,

\begin{align}&{x^{2\left( {k + 1} \right)}} - {y^{2\left( {k + 1} \right)}}\\\Rightarrow \;&{x^{2k + 2}} - {y^{2k + 2}}\\\Rightarrow \;&{x^2}\left( {{x^{2k}}} \right) - {y^2}\left( {{y^{2k}}} \right)\\\Rightarrow\;& {x^2}\left( {{x^{2k}} - {y^{2k}} + {y^{2k}}} \right) - {y^2}\left( {{y^{2k}}} \right)\\\Rightarrow \;&{x^2}\left( {{x^{2k}} - {y^{2k}}} \right) + {x^2}{y^{2k}} - {y^2}\left( {{y^{2k}}} \right)\\\Rightarrow \;&{x^2}.a\left( {x + y} \right) + {y^{2k}}\left( {{x^2} - {y^2}} \right) \qquad \left[ {{\rm{from}}\left( {\rm{1}} \right)} \right]\\\Rightarrow \;&{x^2}.a\left( {x + y} \right) + {y^{2k}}\left( {x + y} \right)\left( {x - y} \right)\\\Rightarrow \;&\left( {x + y} \right)\left[ {a{x^2} + \left( {x - y} \right){y^{2k}}} \right]\end{align}

From the last line, we see that

$$\left( {x + y} \right)\left[ {a{x^2} + \left( {x - y} \right){y^{2k}}} \right]$$is divisible by $$x + y$$.

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 22

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

$${3^{2n + 2}} - 8n - 9$$ is divisible by $$8.$$

### Solution

We can write

$$P\left( n \right):{3^{2n + 2}} - 8n - 9$$is divisible by $$8.$$

We note that

$$P\left( 1 \right):{3^{2.1 + 2}} - 8.1 - 9 = {3^4} - 8 - 9 = 81 - 17 = 64$$ which is divisible by $$8.$$

Thus $$P\left( n \right)$$ is true for $$n = 1$$

Let $$P\left( k \right)$$ be true for some natural number $$k,$$

i.e., $$P\left( k \right):{3^{2k + 2}} - 8k - 9$$ is divisible by $$8.$$

We can write

${3^{2k + 2}} - 8k - 9 = 8a \qquad \quad \ldots \left( 1 \right)$

where $$a \in N$$ .

Now, we will prove that $$P\left( {k + 1} \right)$$ is true whenever $$P\left( k \right)$$ is true.

Now,

\begin{align}&{3^{2\left( {k + 1} \right) + 2}} - 8\left( {k + 1} \right) - 9\\\Rightarrow \;&{3^{2k + 4}} - 8k - 8 - 9\\\Rightarrow \;&{3^2}{.3^{2k + 2}} - 8k - 17\\\Rightarrow \;&{3^2}\left( {{3^{2k + 2}} - 8k - 9 + 8k + 9} \right) - 8k - 17\\\Rightarrow \;&{3^2}\left( {{3^{2k + 2}} - 8k - 9} \right) + {3^2}\left( {8k + 9} \right) - 8k - 17\\\Rightarrow \;&{3^2}.8a + 72k + 81 - 8k - 17 \qquad \quad\left[ {{\rm{from}}\left( {\rm{1}} \right)} \right]\\\Rightarrow \;&9.8a + 64k + 64\\\Rightarrow \;&8\left( {9a + 8k + 8} \right)\end{align}

From the last line, we see that

$$8\left( {9a + 8k + 8} \right)$$ is divisible by $$8.$$

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 23

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

$${41^n} - {14^n}$$ is a multiple of $$27.$$

### Solution

We can write

$$P\left( n \right):{41^n} - {14^n}$$ is a multiple of $$27.$$

We note that

$$P\left( 1 \right):{41^1} - {14^1} = 41 - 14 = 27$$ which is a multiple of $$27.$$

Thus $$P\left( n \right)$$ is true for $$n = 1$$

Let $$P\left( k \right)$$ be true for some natural number $$k,$$

i.e., $$P\left( k \right):{41^k} - {14^k}$$ is a multiple of $$27.$$

We can write

$${41^k} - {14^k} = 27a \qquad \quad \ldots \left( 1 \right)$$

where $$a \in N$$ .

Now, we will prove that $$P\left( {k + 1} \right)$$ is true whenever $$P\left( k \right)$$ is true.

Now,

\begin{align}&{41^{k + 1}} - {14^{k + 1}}\\\Rightarrow \;&{41.41^k} - {14.14^k}\\\Rightarrow \;&41.\left( {{{41}^k} - {{14}^k} + {{14}^k}} \right) - {14.14^k}\\\Rightarrow \;&41.\left( {{{41}^k} - {{14}^k}} \right) + {41.14^k} - {14.14^k}\\\Rightarrow \;&41.27a + {14^k}\left( {41 - 14} \right) \qquad \qquad\left[ {{\rm{from}}\left( {\rm{1}} \right)} \right]\\\Rightarrow \;&41.27a + {14^k}.27\\\Rightarrow \;&27\left( {41a + {{14}^k}} \right)\end{align}

From the last line, we see that

$$27\left( {41a + {{14}^k}} \right)$$ is a multiple of $$27.$$

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Chapter 4 Ex.4.1 Question 24

Prove the following by using the principle of mathematical induction for all $$n \in N$$:

$\left( {2n + 7} \right) < {\left( {n + 3} \right)^2}.$

### Solution

Let $$P\left( n \right)$$ be the given statement.

i.e., $$P\left( n \right):\left( {2n + 7} \right) < {\left( {n + 3} \right)^2}$$

We note that $$P\left( n \right)$$ is true for $$n = 1$$,

Since,

$$P\left( 1 \right):\left( {2.1 + 7} \right) = 9 < {\left( {1 + 3} \right)^2} = 16$$

Assume that $$P\left( k \right)$$ is true for some positive integer $$k$$

i.e., $$\left( {2k + 7} \right) < {\left( {k + 3} \right)^2} \qquad \quad\ldots \left( 1 \right)$$

We will now prove that $$P\left( {k + 1} \right)$$ is true whenever $$P\left( k \right)$$ is true

Now, we have

\begin{align}2\left( {k + 1} \right) + 7 &= 2k + 2 + 7\\2\left( {k + 1} \right) + 7 &= \left( {2k + 7} \right) + 2 < {\left( {k + 3} \right)^2} + 2 \qquad \left[ {{\rm{from}}\;\left( {\rm{1}} \right)} \right]\\\left( {2k + 7} \right) + 2 &< {\left( {k + 3} \right)^2} + 2\\&< {k^2} + 6k + 9 + 2\\&< {k^2} + 6k + 11\end{align}

Now,

\begin{align}{\left[ {\left( {k + 1} \right) + 3} \right]^2} &= {\left( {k + 4} \right)^2}\\&= {k^2} + 8k + 16\end{align}

Since,

${k^2} + 6k + 11 < {k^2} + 8k + 16$

Therefore,

\begin{align}&2\left( {k + 1} \right) + 7 < {\left( {k + 4} \right)^2}\\&\left[ {2\left( {k + 1} \right) + 7} \right] < {\left[ {\left( {k + 1} \right) + 3} \right]^2}\end{align}

Thus $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.

Hence, from the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for all natural numbers i.e., $$n \in N$$.

## Exercise 4.1

Download FREE PDF of Chapter-4 Principle of Mathematical Induction
Principle of Mathematical Induction | NCERT Solutions

### math teachers and top

Personalized Curriculum
Instant Doubts clarification
Cover latest CBSE Syllabus
Unlimited Mock & Practice tests
Covers CBSE, ICSE, IB curriculum

Instant doubt clearing with Cuemath Advanced Math Program
Â Â
Related Sections
Related Sections