NCERT Solutions For Class 11 Maths Chapter 9 Exercise 9.4

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Chapter 9 Ex.9.4 Question 1

Find the sum to \(n\) terms of the series \(1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + \ldots \)

 

Solution

 

The given series is \(1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + \ldots n{\rm{ terms}}\)

Hence,

\({a_n} = n\left( {n + 1} \right)\)

Therefore,

\[\begin{align}{S_n} &= \sum\limits_{k = 1}^n {{a_k}}  = \sum\limits_{k = 1}^n {k\left( {k + 1} \right)} \\ &= \sum\limits_{k = 1}^n {{k^2} + } \sum\limits_{k = 1}^n k \\ &= \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \frac{{n\left( {n + 1} \right)}}{2}\\& = \frac{{n\left( {n + 1} \right)}}{2}\left( {\frac{{2n + 1}}{3} + 1} \right)\\ &= \frac{{n\left( {n + 1} \right)}}{2}\left( {\frac{{2n + 4}}{3}} \right)\\ &= \frac{n}{3}\left( {n + 1} \right)\left( {n + 2} \right)\end{align}\]

Chapter 9 Ex.9.4 Question 2

Find the sum to terms of the series \(1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + 4 \times 5 + \ldots \)

   

Solution

The given series is \(1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + 4 \times 5 + \ldots n{\rm{ terms}}\)

Hence,

\[\begin{align}{a_n} &= n\left( {n + 1} \right)\left( {n + 2} \right)\\&= \left( {{n^2} + n} \right)\left( {n + 2} \right)\\&= {n^3} + 3{n^2} + 2n\end{align}\]

Therefore,

\[\begin{align}{S_n} &= \sum\limits_{k = 1}^n {{a_k}} \\&= \sum\limits_{k = 1}^n {\left( {{k^3} + 3{k^2} + 2k} \right)} \\&= \sum\limits_{k = 1}^n {{k^3} + 3} \sum\limits_{k = 1}^n {{k^2} + 2} \sum\limits_{k = 1}^n k \\&= {\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2} + \frac{{3n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \frac{{2n\left( {n + 1} \right)}}{2}\\&= {\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2} + \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{2} + n\left( {n + 1} \right)\end{align}\]

\[\begin{align}{S_n} &= \frac{{n\left( {n + 1} \right)}}{2}\left[ {\frac{{n\left( {n + 1} \right)}}{2} + 2n + 1 + 2} \right]\\&= \frac{{n\left( {n + 1} \right)}}{2}\left[ {\frac{{{n^2} + n + 4n + 6}}{2}} \right]\\&= \frac{{n\left( {n + 1} \right)}}{4}\left[ {{n^2} + 5n + 6} \right]\\&= \frac{{n\left( {n + 1} \right)}}{4}\left[ {{n^2} + 2n + 3n + 6} \right]\\&= \frac{{n\left( {n + 1} \right)\left[ {n\left( {n + 2} \right) + 3\left( {n + 2} \right)} \right]}}{4}\\&= \frac{n}{4}\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)\end{align}\]

Chapter 9 Ex.9.4 Question 3

Find the sum to \(n\) terms of the series \(3 \times {1^2} + 5 \times {2^2} + 7 \times {3^2} + \ldots \)

  

Solution

The given series is \(3 \times {1^2} + 5 \times {2^2} + 7 \times {3^2} + \ldots n{\rm{ terms}}\)

Hence,

\[\begin{align}{a_n}&= \left( {2n + 1} \right){n^2}\\&= 2{n^3} + {n^2}\end{align}\]

Therefore,

\[\begin{align}{S_n} &= \sum\limits_{k = 1}^n {{a_k}} \\&= \sum\limits_{k = 1}^n {\left( {2{k^3} + {k^2}} \right)} \\&= 2\sum\limits_{k = 1}^n {{k^3} + } \sum\limits_{k = 1}^n {{k^2}} \\&= 2{\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2} + \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\\&= \frac{{{n^2}{{\left( {n + 1} \right)}^2}}}{2} + \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\\&= \frac{{n\left( {n + 1} \right)}}{2}\left[ {n\left( {n + 1} \right) + \frac{{2n + 1}}{3}} \right]\\&= \frac{{n\left( {n + 1} \right)}}{2}\left[ {\frac{{3{n^2} + 3n + 2n + 1}}{3}} \right]\\&= \frac{{n\left( {n + 1} \right)}}{2}\left[ {\frac{{3{n^2} + 5n + 1}}{3}} \right]\\&= \frac{n}{6}\left( {n + 1} \right)\left( {3{n^2} + 5n + 1} \right)\end{align}\]

Chapter 9 Ex.9.4 Question 4

Find the sum to \(n\)terms of the series \(\frac{1}{{1 \times 2}} + \frac{1}{{2 \times 3}} + \frac{1}{{3 \times 4}} + \ldots \)

 

Solution

 

The given series is \(\frac{1}{{1 \times 2}} + \frac{1}{{2 \times 3}} + \frac{1}{{3 \times 4}} + \ldots n{\rm{ terms}}\)

Hence,

\[\begin{align}{a_n}&= \frac{1}{{n\left( {n + 1} \right)}}\\&= \frac{1}{n} - \frac{1}{{n + 1}}\end{align}\]

Therefore,

\[\begin{align}{a_1} &= \frac{1}{1} - \frac{1}{2}\\{a_2} &= \frac{1}{2} - \frac{1}{3}\\{a_3}& = \frac{1}{3} - \frac{1}{4}\\{a_n} &= \frac{1}{n} - \frac{1}{{n + 1}}\end{align}\]

Adding the above terms column wise, we obtain

\[{a_1} + {a_2} + { \ldots _n} = \left[ {\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + . \ldots \frac{1}{n}} \right] - \left[ {\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots \frac{1}{{n + 1}}} \right]\]

Thus,

\[\begin{align}{S_n} &= 1 - \frac{1}{{n + 1}}\\&= \frac{{n + 1 - 1}}{{n + 1}}\\&= \frac{n}{{n + 1}}\end{align}\]

Chapter 9 Ex.9.4 Question 5

Find the sum to \(n\) terms of the series \({5^2} + {6^2} + {7^2} + \cdots + {20^2}\)

Solution

The given series is \({5^2} + {6^2} + {7^2} + \cdots + {20^2}\)

Hence,

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

Therefore,

\[\begin{align}{S_n}&= \sum\limits_{k = 1}^n {{a_k}} \\&= \sum\limits_{k = 1}^n {\left( {{k^2} + 8k + 16} \right)} \end{align}\]

\[\begin{align}{S_n} &= \sum\limits_{k = 1}^n {{k^2} + 8} \sum\limits_{k = 1}^n k + \sum\limits_{k = 1}^n {16} \\&= \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \frac{{8n\left( {n + 1} \right)}}{2} + 16n\end{align}\]

Since, \(16^{th}\) term is \({\left( {16 + 4} \right)^2} = {\left( {20} \right)^2}\)

Then,

\[\begin{align}{S_{16}} &= \frac{{16\left( {16 + 1} \right)\left( {2\left( {16} \right) + 1} \right)}}{6} + \frac{{8\left( {16} \right)\left( {16 + 1} \right)}}{2} + 16\left( {16} \right)\\&= \frac{{\left( {16} \right)\left( {17} \right)\left( {33} \right)}}{6} + \frac{{\left( {128} \right)\left( {17} \right)}}{2} + 256\\&= 1496 + 1088 + 256\\&= 2840\end{align}\]

Thus, \({5^2} + {6^2} + {7^2} + \cdots + {20^2} = 2840\)

Chapter 9 Ex.9.4 Question 6

Find the sum to \(n\)terms of the series \(3 \times 8 + 6 \times 11 + 9 \times 14 + \ldots \)

Solution

The given series is \(3 \times 8 + 6 \times 11 + 9 \times 14 + \ldots n{\rm{ terms}}\)

Hence,

\[\begin{align}{a_n} &= \left( {{n^{th}}{\text{ term of }}3,6,9 \ldots } \right) \times \left( {{n^{th}}{\text{ term of }}8,11,14 \ldots } \right)\\&= \left( {3n} \right)\left( {3n + 5} \right)\\&= 9{n^2} + 15n\end{align}\]

Therefore,

\[\begin{align}{S_n}& = \sum\limits_{k = 1}^n {{a_k}} \\&= \sum\limits_{k = 1}^n {\left( {9{k^2} + 15k} \right)} \\&= 9\sum\limits_{k = 1}^n {{k^2} + 15} \sum\limits_{k = 1}^n k \\&= 9 \times \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + 15 \times \frac{{n\left( {n + 1} \right)}}{2}\\&= \frac{{3n\left( {n + 1} \right)\left( {2n + 1} \right)}}{2} + \frac{{15n\left( {n + 1} \right)}}{2}\\&= \frac{{3n\left( {n + 1} \right)}}{2}\left( {2n + 1 + 5} \right)\\&= \frac{{3n\left( {n + 1} \right)}}{2}\left( {2n + 6} \right)\\&= 3n\left( {n + 1} \right)\left( {n + 3} \right)\end{align}\]

Chapter 9 Ex.9.4 Question 7

Find the sum to terms of the series \({{1}^{2}}+\left( {{1}^{2}}+{{2}^{2}} \right)+\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}} \right)+\ldots \)

   

Solution

   

The given series is \({{1}^{2}}+\left( {{1}^{2}}+{{2}^{2}} \right)+\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}} \right)+\ldots n\text{ terms}\)

Hence,

\[\begin{align}  {{a}_{n}}&=\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+\ldots +{{n}^{2}} \right) \\ & =\frac{n}{6}\left( n+1 \right)\left( 2n+1 \right) \\ & =\frac{n}{6}\left( 2{{n}^{2}}+3n+1 \right) \\ & =\frac{1}{3}{{n}^{3}}+\frac{1}{2}{{n}^{2}}+\frac{1}{6}n \end{align}\]

Therefore,

\[\begin{align}{S_n} &= \sum\limits_{k = 1}^n {{a_k}} \\&= \sum\limits_{k = 1}^n {\left( {\frac{1}{3}{k^3} + \frac{1}{2}{k^2} + \frac{1}{6}k} \right)} \\&= \frac{1}{3}\sum\limits_{k = 1}^n {{k^3} + } \frac{1}{2}\sum\limits_{k = 1}^n {{k^2}} + \frac{1}{6}\sum\limits_{k = 1}^n k \\&= \frac{1}{3} \times \frac{{{n^2}{{\left( {n + 1} \right)}^2}}}{{{{\left( 2 \right)}^2}}} + \frac{1}{2} \times \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \frac{1}{6} \times \frac{{n\left( {n + 1} \right)}}{2}\\&= \frac{{n\left( {n + 1} \right)}}{6}\left[ {\frac{{n\left( {n + 1} \right)}}{2} + \frac{{\left( {2n + 1} \right)}}{2} + \frac{1}{2}} \right]\\&= \frac{{n\left( {n + 1} \right)}}{6}\left[ {\frac{{{n^2} + n + 2n + 1 + 1}}{2}} \right]\\&= \frac{{n\left( {n + 1} \right)}}{6}\left[ {\frac{{{n^2} + n + 2n + 2}}{2}} \right]\\&= \frac{{n\left( {n + 1} \right)}}{6}\left[ {\frac{{n\left( {n + 1} \right) + 2\left( {n + 1} \right)}}{2}} \right]\\&= \frac{{n\left( {n + 1} \right)}}{6}\left[ {\frac{{\left( {n + 1} \right)\left( {n + 2} \right)}}{2}} \right]\\&= \frac{{n{{\left( {n + 1} \right)}^2}\left( {n + 2} \right)}}{{12}}\end{align}\]

Chapter 9 Ex.9.4 Question 8

Find the sum to \(n\) terms of the series whose \({n^{th}}\) term is given by \(n\left( {n + 1} \right)\left( {n + 4} \right)\)

Solution

The given \({n^{th}}\) term is \({a_n} = n\left( {n + 1} \right)\left( {n + 4} \right)\)

Hence,

\[\begin{align}{a_n}&= n\left( {n + 1} \right)\left( {n + 4} \right)\\&= n\left( {{n^2} + 5n + 4} \right)\\&= {n^3} + 5{n^2} + 4n\end{align}\]

Therefore,

\[\begin{align}
{S_n} &= \sum\limits_{k = 1}^n {{a_k}} \\
 &= \sum\limits_{k = 1}^n {{k^3} + 5\sum\limits_{k = 1}^n {{k^2}}  + 4\sum\limits_{k = 1}^n k } \\
& = \frac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4} + \frac{{5n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \frac{{4n\left( {n + 1} \right)}}{2}\\
& = \frac{{n\left( {n + 1} \right)}}{2}\left[ {\frac{{n\left( {n + 1} \right)}}{2} + \frac{{5\left( {2n + 1} \right)}}{3} + 4} \right]\\
 &= \frac{{n\left( {n + 1} \right)}}{2}\left[ {\frac{{3{n^2} + 3n + 20n + 10 + 24}}{6}} \right]\\
 &= \frac{{n\left( {n + 1} \right)}}{2}\left[ {\frac{{3{n^2} + 23n + 34}}{6}} \right]\\
& = \frac{{n\left( {n + 1} \right)\left( {3{n^2} + 23n + 34} \right)}}{{12}}
\end{align}\]

Chapter 9 Ex.9.4 Question 9

Find the sum to \(n\) terms of the series whose \({{n}^{th}}\) term is given by \({{n}^{2}}+{{2}^{n}}\).

Solution

The given \({{n}^{th}}\) term is \({a_n} = {n^2} + {2^n}\)

Hence

\[\begin{align}{S_n} &= \sum\limits_{k = 1}^n {{k^2} + {2^k}} \\ &= \sum\limits_{k = 1}^n {{k^2}}  + \sum\limits_{k = 1}^n {{2^k}} \;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)
\end{align}\]

Consider \(\sum\limits_{k = 1}^n {{2^k}} = {2^1} + {2^2} + {2^3} + \ldots \)

The above series \({2^1} + {2^2} + {2^3} +  \ldots \) is a G.P with both the first term and common ratio equal to \(2\).

\[\begin{align}\sum\limits_{k = 1}^n {{2^k}} &= \frac{{\left( 2 \right)\left[ {{{\left( 2 \right)}^n} - 1} \right]}}{{2 - 1}}\\&= 2\left( {{2^n} - 1} \right)\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}\]

From and \(\left( 2 \right)\), we obtain

\[\begin{align} {{S}_{n}}&=\sum\limits_{k=1}^{n}{{{k}^{2}}}+2\left( {{2}^{n}}-1 \right) \\ & =\frac{n}{6}\left( n+1 \right)\left( 2n+1 \right)+2\left( {{2}^{n}}-1 \right) \end{align}\]

Chapter 9 Ex.9.4 Question 10

Find the sum to \(n\) terms of the series whose \({n^{th}}\) term is given by \({\left( {2n - 1} \right)^2}\) .

Solution

The given \({n^{th}}\) term is \({a_n} = {\left( {2n - 1} \right)^2}\)

Hence,

\[\begin{align}{a_n} &= {\left( {2n - 1} \right)^2}\\&= 4{n^2} - 4n + 1\end{align}\]

Therefore,

\[\begin{align}{S_n} &= \sum\limits_{k = 1}^n {{a_k}} \\&= \sum\limits_{k = 1}^n {\left( {4{k^2} - 4k + 1} \right)} \\&= 4\sum\limits_{k = 1}^n {{k^2}} - 4\sum\limits_{k = 1}^n k + \sum\limits_{k = 1}^n 1 \\&= \frac{{4n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \frac{{4n\left( {n + 1} \right)}}{2} + n\\&= \frac{{2n\left( {n + 1} \right)\left( {2n + 1} \right)}}{3} - 2n\left( {n + 1} \right) + n\\&= n\left[ {\frac{{2\left( {2{n^2} + 3n + 1} \right)}}{3} - 2\left( {n + 1} \right) + 1} \right]\\&= n\left[ {\frac{{4{n^2} + 6n + 2 - 6n - 6 + 3}}{3}} \right]\\&= n\left[ {\frac{{4{n^2} - 1}}{3}} \right]\\&= \frac{n}{3}\left( {2n + 1} \right)\left( {2n - 1} \right)\end{align}\]