Step1: For n=1, P(1):

LHS=S₁=1

RHS=S₁=1

So, P(1) is true.

Step2: Let P(n) be true for n=k

Now, we need to show P(k+1) is true whenever P(k) is true.

P(k+1):

Case1: When k is odd, then (k+1) is even

As LHS=RHS

So, it is true for n=k+1 when k is odd.

Case2: When k is even, then (k+1) is odd

As LHS=RHS

So, it is true for n=k+1 when k is even.

Hence, by the principle of mathematical induction P(n) is true ∀ nϵN.