In mathematical induction, we find whether a given statement is true for all the natural numbers or not. It is basically a technique used to prove a statement or a theorem, or a formula that is advanced about all natural numbers. The simplest form of mathematical induction is to identify that whether a statement which include a natural number 'n' holds for all values of n is true or not. |

- P(1) is true.
- P(k + 1) is true, whenever P(k) is true.

This method of proof of a proposition P(n) by mathematical induction consists of the following steps:

**Verify the proposition P(n) for n = 1.**

__Basic step:__**Assume the proposition P(n) to be true for some particular value of n say n = k, and prove that the result is true for n = k + 1. That is, show that P(k + 1) is true.**

__Induction step:__**This completes the principle of induction. Conclude by saying, "By principle of mathematical induction, P(n) is true for all natural number 'n'".**

__Conclusion step:__Second Principle of Mathematical Induction is a strong mathematical induction. Let j be an integer, and let T(n) be a predicate whose universe of discourse is the set of integers {j, j + 1, j + 2, ....}.

Assume that

- T(j), and
- T(j) for j $\leq $ i $\leq$ n implies P(n + 1). Then, T(n) is true $\forall$ n.

The second principle of induction is also known as strong mathematical induction. → Read More The proof of Mathematical Induction contains two steps, and the steps are shown below:

- The basic case of induction mathematical is used to show that the statement holds the value of n is equal to the lowest value of n. Generally, we take the value of n as 0 or 1.
- The inductive case of mathematical induction is used
to show that the given statement holds the value of n. Then, the full
statement also holds, when the value of n is substituted to n + 1.

The steps for mathematical induction are as follows:

- For any given theorem, first we have to show that the given condition is true for n = 1.
- Assume that it is true for n = k.
- Explain that result is true for n = k + 1
**Conclusion:**Given statement is true for all values of n $\geq$ 1.

### Solved Example

**Question:**Prove by the principle of mathematical induction $n^{2} \geq 2n$, where n = 2, 3, 4 and so on.

**Solution:**

**Step 1:**Let n = 2, L.H.S = $2^{2}$ = 4,

R.H.S = 2 x 2 =4

$\therefore$ L.H.S = R.H.S $\Rightarrow$ P(2) is true.

**Step 2:**Let P(n) be true for n = k.

$k^{2} \geq 2k$ -------1

**Step 3:**We shall show that P(k + 1) is true.

Let n = k + 1. Then, $(k + 1)^{2}$ $\geq 2(k + 1)$ for every (k + 1) = 2, 3, 4 and so on.

$(k + 1) ^{2} = k^{2} + 1 + 2k$

$\geq 2k + 1 + 2k$ (Using step 1)

= $2k+ (1 + 2k)$

$\geq 2k + 2$

Thus, we have $(k + 1)^{2} \geq 2(k + 1)$

This statement is same as P(n), when n = k + 1

This shows that P(n) is true for n = k + 1.

Thus, P(k + 1) is true whenever P(k) is true.

**Step 4:**By principle of mathematical induction, P(n) is true for all natural number 'n'.

### Solved Examples

**Question 1:**Prove by the principle of mathematical induction

$1^{2}$ +$2^{2}$ +.........+ $n^{2}$ = $\frac{n(n+1)(2n+1)}{6}$ for all positive integer n.

**Solution:**

**Let n = 1, L.H.S = 1$^{2}$ = 1, R.H.S = $\frac{1(1+1)(2+1)}{6}$**

Step 1:

Step 1:

=$\frac{6}{6}$ = 1

$\therefore$ L.H.S = R.H.S $\Rightarrow$ P(1) is true.

**Step 2:**Let P(n) be true for n = k.

$1^{2}$ + $2^{2}$ +.........+$k^{2}$ = $\frac{k(k+1)(2k+1)}{6}$ ----1

**Step 3:**we shall show that P (k + 1) is true.

Adding both sides with (k+1)$^{2}$, we get

$1^{2}$ + $2^{2}$ +.........+$k^{2}$ + $(k+1)^{2}$

=$\frac{k(k+1)(2k+1)}{6}$$ + (k + 1)^2$

=$\frac{k(k+1)(2k+1)+6(k+1)^{2}}{6}$

= $\frac{(k+1)[k(2k+1)+6(k+1)]}{6}$

=$\frac{(k+1)[2k^{2}+7k+6]}{6}$

=$\frac{(k+1)(k+2)(2k+3)}{6}$

=$\frac{(k+1)(k+2)[2(k+1)+1]}{6}$

Thus, we have, $1^{2}$ +$2^{2}$ +.........+$k^{2}$+ $(k+1)^{2}$ = $\frac{(k+1)(k+2)[2(k+1)+1]}{6}$

This statement is same as P(n), when n = k + 1

This shows that P(n) is true for n = k + 1.

Thus, P(k + 1) is true, whenever P(k) is true.

**Step 4:**By principle of mathematical induction, P(n) is true for all natural number 'n'.

**Question 2:**$\frac{1}{1.2}$ + $\frac{1}{2.3}$ + $\frac{1}{3.4}$ +.....+ $\frac{1}{n.(n+1)}$ =$\frac{n}{n+1}$ for all positive integer n.

**Solution:**

P(n) = $\frac{1}{1.2}$ + $\frac{1}{2.3}$ + $\frac{1}{3.4}$+.......+ $\frac{1}{n.(n+1)}$=$\frac{n}{n+1}$

**Step 1:**Let n = 1, L.H.S = $\frac{1}{1. 2}$ = $\frac{1}{2}$

R.H.S = $\frac{1}{1+1}$ = $\frac{1}{2}$

L.H.S = R.H.S $\Rightarrow$ P(1)is true.

**Step 2:**Let P(n) be true for n = k

$\frac{1}{1.2}$ + $\frac{1}{2.3}$ + $\frac{1}{3.4}$ +.....+$\frac{1}{k(k+1)}$ = $\frac{k}{k+1}$ ....1

**Step 3:**We shall show that P(n) is true for n = k + 1.

Adding both sides of (1) with $\frac{1}{(k+1)(k+2)}$, we get

$\frac{1}{1.2}$ + $\frac{1}{2.3}$ +$\frac{1}{3.4}$ +.....+$\frac{1}{k(k+1)}$ +$\frac{1}{(k+1)(k+2)}$ = $\frac{k}{k+1}$ +$\frac{1}{(k+1)(k+2)}$

= $\frac{k(k+2)+1}{(k+1)(k+2)}$

= $\frac{k^{2}+2k+1}{(k+1)(k+2)}$

= $\frac{(k+1)^{2}}{(k+1)(k+2)}$

= $\frac{(k+1)}{(k+2)}$

Thus, we have

$\frac{1}{1.2}$ + $\frac{1}{2.3}$ +$\frac{1}{3.4}$ +.....+$\frac{1}{k(k+1)}$ +$\frac{1}{(k+1)(k+2)}$ = $\frac{(k+1)}{(k+2)}$

This statement is same as P(n), when n = k + 1.

This shows that P(n) is true for n = k + 1.

Thus, P(k + 1) is true whenever P(k) is true.

**Step 4:**By principle of mathematical induction, P(n) is true for all natural number 'n'.