Question

Prove by the Principle of Mathematical Induction: \(3^2n−1\) is divisible by 8, for all natural numbers n.

Answer 1

Assume given statement

Let \(P(n)=3^{2n}−1\) is divisible by 8 \(\forall n\epsilon \mathbb{N}\).

Check that statement is true for \(n=1\)

\(\because P(1)=3^{2(1)}−1=3^2−1=9−1=8\)

\(\Rightarrow P(1) \)is divisible by 8

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

Assume P(k) to be true and then prove \(P(k+1)\) is true.

Assume that P(k) is divisible by 8 for all \(k\epsilon\mathbb{N}\)

So, P(k) is true i.e.,\(3^{2k}−1=8q\)

\(\Rightarrow 3^{2k}=8q+1,\) where \(q\epsilon\mathbb{N}…(1)\)

\(P(k+1)=3^{2(k+1)}−1\)

\(=3^{2k+2}−1\)

\(=3^{2k}.3^2−1\)

\(=9.3^{2k}−1 P(k+1)\)

\(=9(8q+1)−1\)

\(=72q+9−1 =8(9q+1),\) where \(q\epsilon \mathbb{N}.\)

So, \(P(k+1)\) is divisible by 8

Hence, \(P(k+1)\) is true whenever P(k) is true.

Hence, By Principle of mathematical Induction P(n) is true for all natural numbers n.