In Mathematics, we see many theorems which are true for all natural numbers. Sometimes, proving these theorems directly is not possible. For this purpose, we use a method of proof which is known as “Mathematical Induction”.
The principal of Mathematical Induction is defined as follows.
“Let S(n) be a certain statement involving n, where n is a natural number.
If S(1) is true and if S(k) implies S(k+1) where k is a natural number, then S(n) is true for all natural numbers n.”
Explanation:
The principle of mathematical induction can be explained in simple words as follows.
Step 1: Prove S(1).
Step 2: Assume that S(k) is true doe some natural number k.
Step 3: Prove S(k+1).
From step-1, S(n) is true for n=1. From step-3, S(n) is true for n=k+1, then automatically S(n) is true for n=1+1=2.
Also, S(n) is true for n=2+1=3.
Similarly S(n) is true for n=4,5,…..
So, by this principle of Mathematical induction, we can say that S(n) is true for all natural numbers.
Example 1:
By the principle of mathematical induction, prove that
1+2+3+…+n = n(n+1)/2.
Solution:
Let S(n) : 1+2+3+…+n = n(n+1)/2.
Step 1: Prove S(1).
When n=1,
LHS = 1, RHS = 1(1+1)/2=1.
LHS=RHS
So, S(1) is proved.
Step 2: Assume that S(k) is true doe some natural number k.
1+2+3+…+k = k(k+1)/2.
Step 3: Prove S(k+1).
LHS=1+2+3+…+(k+1)
=(1+2+3+…+k) + (k+1)
= k(k+1)/2 + (k+1)
= [k(k+1) + 2(k+1)]/2
=(k+1)(k+2)/2
=(k+1)[(k+1)+1]/2
=RHS.
So, S(k+1) is true.
So, by this principle of Mathematical induction, we can say that S(n) is true for all natural numbers.
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