## Principle of Mathematical Induction:-

we know that the first positive even integer is 2 = 2 1

the second positive even integer is 4 =

the third positive even integer is 6 = 23

the fourth positive even integer is 8 = 24

………………………………………………….

If we move using the same pattern then nth positive integer=2n

but this is just the summary of the above observations and it’s a statement which we believe is true. We will call these types of statements “**proposition”. **A proposition will remain a proposition until we prove it true. We represent a proposition by a symbol .

**PRINCIPLE OF MATHEMATICAL INDUCTION(PMI)-** “Principle of Mathematical induction proves that we can climb on a ladder as high as we wish, by proving that we can climb the first rung/bottom rung (**the base**) and from each rung, we can go up to the next rung”

If we assume a mathematical **proposition()** to be like a ladder we can say that

(i) If is true for the first term where (*this is called base step*)

(ii) If is true for the (k+1) terms while it’s already true for k terms where (*this is called induction step)*

Then according to the Principle of Mathematical Induction, will be true for all

**Practical Example of PMI (Domino Effect) -:**

Domino effect is one of the best examples of PMI

(i) If the first domino hits the second one then second will hit the next

(ii) If k-th dominos falls the (k+1)th will also fall

This approves the Principle of Mathematical Induction

**Numerical Example of (PMI) :-**

Prove that 1^{ 2} + 2^{ 2} + 3^{ 2} + … + n^{ 2} = is true where

**Solution:-** To prove this statement we use the principle of mathematical induction

- Base Step: We first check if p (1) is true. L.H.S = 1
^{ 2}= 1R.H.S = 1 (1 + 1) (21 + 1)/ 6 = 1 - Both sides are equal so for n=1 it is true.
- Induction Step: We now assume that p (k) is true 1
^{ 2}+ 2^{ 2}+ 3^{ 2}+ … + k^{ 2}= k (k + 1) (2k + 1)/ 6 ……(i) - now we will check for p (k + 1). So we add (k + 1)th term to both sides of the above statement 1
^{ 2}+ 2^{ 2}+ 3^{ 2}+ … + k^{ 2}+ (k + 1)^{ 2}= k (k + 1) (2k + 1)/ 6 + (k + 1)^{ }= (k + 1) [ k (2k + 1)+ 6 (k + 1) ] /6 - now we simplify this = (k + 1) [ 2k
^{ 2}+ 7k + 6 ] /6= (k + 1) [ (k + 2) (2k + 3) ] /6 - We have begun with P(k) and we have proved that1
^{ 2}+ 2^{ 2}+ 3^{ 2}+ … + k^{ 2}+ (k + 1)^{ 2}= (k + 1) [ (k + 2) (2k + 3) ] /6 - Which is the statement P(k + 1).

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