Mia was asked to find the sum of the first 100 natural numbers. She was able to recall this formula:

\[ 1+2+3+...+n = \dfrac{n(n+1)}{2}\]

She substituted \(n=100\) in this formula and got the answer as 5,050

Is this answer correct?

How can you prove this formula?

In this lesson, you will learn all about mathematical induction and learn how to prove this formula. You will learn about the principle of mathematical induction and will go through various mathematical induction examples.

Let's begin!

**Lesson Plan**

**What Is Mathematical Induction?**

Mathematical induction is the process of proving a mathematical statement (or) an equation (or) a theorem for the set of all natural numbers, \(N\).

**Principle of Mathematical Induction**

The principle of mathematical induction works in the following way.

To show something works all the time:

- We have to show that it works for the first time.
- We should assume it works for this time.
- Prove that it works for the next time.

From the above steps, we can then say that it works every time.

This is the principle of mathematical induction in words.

We will see the principle of mathematical induction in mathematical terms in the following section.

**How to Prove Mathematical Induction?**

Using the principle of mathematical induction, to prove a statement \(P(n)\) is true for all \(n\in N\), we have to prove:

- \(P(1)\) is true.
- Assume \(P(k)\) is true for some \(k \in N\).
- Prove \(P(k+1)\) is true.

Here is the proof of the principle of mathematical induction.

**What Are the Steps in Mathematical Induction?**

In the previous section, we have seen the steps to prove a statement by mathematical induction.

Each step is named as follows.

**Base step:**To prove \(P(1)\) is true.**Assumption step:**Assume that \(P(k)\) is true for some \(k \in N\).**Induction step:**Prove that \(P(k+1)\) is true.

After proving these 3 steps, we can say that "By the principle of mathematical induction, \(P(n)\) is true for all \(n \in N\)."

**Example**

Prove the following by mathematical induction:

\[ 1+2+3+...+n = \dfrac{n(n+1)}{2}\]

**Solution**

Let us assume that \[ P(n):\,\,1+2+3+...+n = \dfrac{n(n+1)}{2}\]

We use the above 3 steps to prove \(P(n)\).

**Base step: **Let us prove \(P(n)\) for \(n=1\).

\[\begin{array}{l}

\text { LHS = First term = } 1 \\[0.2cm]

\begin{aligned}

\text{RHS} &=\frac{1(1+1)}{2} \\[0.2cm]

&=\frac{1(2)}{2} \\[0.2cm]

&=1

\end{aligned} \\[0.2cm]

\therefore \quad \text {LHS} = \text {RHS}

\end{array}\]

Thus, \(P(n)\) is true for \(n=1\)

**Assumption step: **Assume that \(P(k)\) is true for some \(k \in N\).

Thus,

\[\begin{align} 1+2+3+...+k &= \dfrac{k(k+1)}{2}\,\,\, \rightarrow (1) \end{align}\]

**Induction step: **Let us prove \(P(k+1)\) is true.

i.e., we have to prove:

\(\begin{align} 1+2+3+&...+(k+1)\\[0.2cm] &= \dfrac{(k+1)[(k+1)+1]}{2}\end{align}\)

\[ \begin{aligned}

\text{LHS} &=1+2+\cdots+(k+1) \\[0.2cm]

&=\underbrace{1+2+\cdots+k}+(k+1) \\[0.2cm]

&=\frac{k(k+1)}{2}+k+1 \quad(\text { From }(1)) \\[0.2cm]

&=\frac{k(k+1)}{2}+\frac{2(k+1)}{2}\\[0.2cm]

&=\frac{k^{2}+k+2 k+2}{2}\\[0.2cm]

&=\frac{k^{2}+3 k+2}{2}\\[0.2cm]

&=\frac{(k+1)(k+2)}{2}\\[0.2cm]

&=\frac{(k+1)((k+1)+1)}{2}\\[0.2cm]

&=\text{RHS}

\end{aligned} \]

Thus, \(P(n)\) is true for \(n=k+1\)

Therefore, by the principle of mathematical induction, \(P(n)\) is true for all \(n \in N\).

**The "induction step" is the most difficult part of the "proof by mathematical induction" method. Here are some tips to get through it:**

**\(k^{th}\) term comes just before \((k+1)^{th}\) term.****The result of the "assumption step" is used after writing the \(k^{th}\) term (before the \((k+1)^{th}\) term).****If getting the RHS from the LHS seems difficult, simplify the LHS and the RHS separately and prove that they are equal.**

**Solved Examples**

Here are a few mathematical induction examples.

Example 1 |

Henry is asked to prove the following by the principle of mathematical induction for all \(n \in N\). Can we help him?

\[\begin{align}1^{2}+3^{2}+5^{2}+&\ldots+(2 n-1)^{2}\\[0.2cm]&=\frac{n(2 n-1)(2 n+1)}{3}\end{align}\]

**Solution**

Let us assume the given equation to be \(P(n)\).

**Base step: **Let us prove \(P(n)\) for \(n=1\)

\[\begin{array}{l}

\text { LHS = First term = } 1^2=1 \\[0.2cm]

\begin{aligned}

\text{RHS} &=\frac{1(2-1)(2+1)}{3} \\[0.2cm]

&=\frac{1(3)}{3} \\[0.2cm]

&=1

\end{aligned} \\[0.2cm]

\therefore \quad \text {LHS} = \text {RHS}

\end{array}\]

Thus, \(P(n)\) is true for \(n=1\)

**Assumption step: **Assume that \(P(k)\) is true for some \(k \in N\). Then we get:

\[\begin{align}1^{2}+&3^{2}+5^{2}+\ldots+(2 k-1)^{2}\\[0.2cm]&=\frac{k(2 k-1)(2 k+1)}{3} \,\, \rightarrow (1) \end{align}\]

**Induction step: **Let us prove \(P(k+1)\) is true.

i.e., we have to prove:

\(\begin{align}&1^{2}+3^{2}+5^{2}+\ldots+[2 (k+1)-1]^{2}\\[0.2cm]&=\dfrac{(k+1)[2(k+1)-1][2(k+1)+1]}{3}\end{align}\)

\( \begin{aligned}

&\text{LHS}\\[0.2cm]

&=1^{2}+3^{2}+5^{2}+\ldots+[2(k+1)-1]^{2}\\[0.2cm]

&=\underbrace{1^{2}+3^{2}+5^{2}+\ldots+(2 k-1)^{2}}+[2 (k+1)-1]^{2}\\[0.2cm]

&=\frac{k(2 k-1)(2 k+1)}{3}+(2 k+1)^{2} \,[\text { from (1)} ]\\[0.2cm]

&=\frac{k(2 k-1)(2 k+1)+3(2 k+1)^{2}}{3}\\[0.2cm]

&=\frac{(2 k+1)[k(2 k-1)+3(2 k+1)]}{3}\\[0.2cm]

&=\frac{(2 k+1)\left[2 k^{2}-k+6 k+3\right]}{3}\\[0.2cm]

&=\frac{(2 k+1)\left[2 k^{2}+5 k+3\right]}{3}\\[0.2cm]

&=\frac{(2 k+1)\left[2 k^{2}+2 k+3 k+3\right]}{3}\\[0.2cm]

&=\frac{(2 k+1)[2 k(k+1)+3(k+1)]}{3}\\[0.2cm]

&=\frac{(2 k+1)(k+1)(2 k+3)}{3}\\[0.2cm]

&=\frac{(k+1)[2(k+1)-1][2(k+1)+1]}{3}\\[0.2cm]

&=\text{RHS}

\end{aligned} \)

Hence, \(P(n)\) is true for \(n=k+1\)

Thus, by the principle of mathematical induction, \(P(n)\) is true for all \(n \in N\).

\(\therefore\) Hence proved |

Example 2 |

Can we help Jake prove the following by the principle of mathematical induction for all \(n \in N\)?

\[\begin{align}a+a r+a r^{2}&+\ldots+a r^{n-1}\\[0.2cm]&=\frac{a\left(r^{n}-1\right)}{r-1}\end{align}\]

**Solution**

Let us assume the given equation to be \(P(n)\).

**Base step: **Let us prove \(P(n)\) for \(n=1\).

\[\begin{array}{l}

\text { LHS =First term = } a \\[0.2cm]

\begin{aligned}

\text{RHS} &=\frac{a\left(r-1\right)}{r-1} \\[0.2cm]

&=a

\end{aligned} \\[0.2cm]

\therefore \quad \text {LHS} = \text {RHS}

\end{array}\]

So \(P(n)\) is true for \(n=1\).

**Assummtion step: **Assume that \(P(k)\) is true for some \(k \in N\). Then we get:

\[\begin{aligned}a+a r&+a r^{2}+\ldots+a r^{k-1}\\[0.2cm]&=\frac{a\left(r^{k}-1\right)}{r-1}\,\rightarrow(1) \end{aligned}\]

**Induction step: **Let us prove \(P(k+1)\) is true.

i.e., we have to prove:

\(\begin{align}a+a r&+a r^{2}+\ldots+a r^{(k+1)-1}\\[0.2cm]& =\frac{a\left(r^{k+1}-1\right)}{r-1}\end{align}\)

\( \begin{aligned}

&\text{LHS}\\[0.2cm]

&=a+a r+a r^{2}+\ldots+a r^{(k+1)-1}\\[0.2cm]

&=\underbrace{\left[a+a r+a r^{2}+\ldots+a r^{k-1}\right]}+a r^{(k+1)-1}\\[0.2cm]

&=\frac{a\left(r^{k}-1\right)}{r-1}+a r^{k} \,\,\,[\text { from (1)} ]\\[0.2cm]

&=\frac{a\left(r^{k}-1\right)+a r^{k}(r-1)}{r-1}\\[0.2cm]

&=\frac{a r^{k}-a+a r^{k+1}-a r^{k}}{r-1}\\[0.2cm]

&=\frac{a r^{k+1}-a}{r-1}\\[0.2cm]

&=\frac{a\left(r^{k+1}-1\right)}{r-1}\\[0.2cm]

&=\text{RHS}

\end{aligned} \)

Hence, \(P(n)\) is true for \(n=k+1\).

Thus, by the principle of mathematical induction, \(P(n)\) is true for all \(n \in N\).

\(\therefore\) Hence proved. |

Example 3 |

Can we help John prove the following by the principle of mathematical induction:

\(10^{2 n-1}+1\) is divisible by 11

**Solution**

Let us assume the given statement to be \(P(n)\).

**Base step: **When \(n = 1\),

\(10^{2 (1)-1}+1 =11\), which is divisible by 11

Hence, \(P(n)\) is true for \(n=1\)

**Assumption step: **Assume that \(P(k)\) is true for some \(k \in N\).

We get:

\(10^{2 k-1}+1\) is divisible by 11

We can write \(10^{2 k-1}+1\) as a multiple of 11

\[10^{2 k-1}+1 = 11a \,\,\,\, \rightarrow (1)\]

Here, \(a\) is any integer.

**Induction step: **Let us prove \(P(k+1)\) is true. For this consider,

\[ \begin{aligned}

&10^{2(k+1)-1}+1 \\[0.2cm]

&=10^{2k+2-1}+1 \\[0.2cm]

&=10^{2 k+1}+1 \\[0.2cm]

&=10^{2}\left(10^{2 k-1}\right)+1 \\[0.2cm]

&=10^{2}\left(10^{2 k-1}+1-1\right)+1 \\[0.2cm]

&=10^{2}\underbrace{\left(10^{2 k-1}+1\right)}-10^{2}+1 \\[0.2cm]

&=10^{2} \cdot 11 a-100+1 \,\,\, \text{ [ From (1)] } \\[0.2cm]

&=10^{2} \cdot 11 a-99 \\[0.2cm]

&=11(100 a-9)

\end{aligned} \]

So \(10^{2(k+1)-1}+1\) is divisible by 11 (as it is a multiple of 11)

So \(P(n)\) is true for \(n=k+1\)

Thus, by the principle of mathematical induction, \(P(n)\) is true for all \(n \in N\).

\(\therefore\) Hence proved |

Prove the following by the principle of mathematical induction:

- \(41^{n}-14^{n}\) is a multiple of 27
- \((2 n+7)<(n+3)^{2}\).

**Interactive Questions**

**Here are a few activities for you to practice. **

**Select/type your answer and click the "Check Answer" button to see the result.**

**Let's Summarize**

We hope you enjoyed learning about Mathematical Induction with the simulations and practice questions. Now, you will be able to easily solve problems on the principle of mathematical induction and mathematical induction examples.

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**Frequently Asked Questions (FAQs)**

## 1. When can you use mathematical induction?

We use mathematical induction to prove a mathematical equation, statement, (or) a theorem.

## 2. Why is mathematical induction important?

Mathematical induction is important because we can use it to prove a mathematical equation, statement, (or) a theorem.

## 3. What is the use of mathematical induction in real life?

Recall the sinking of the Titanic. The crew of the Titanic realized that when a given bulkhead was completely flooded, the next bulkhead would undergo the same fate, thus sinking the whole ship.