Prove that 3

^{n}-1 is divisible by 2 \(\displaystyle \forall n \in\mathbb{Z}^+ ,n\geq 1\)

S(n)=3

^{n}-1=2k, \(\displaystyle k\in\mathbb{Z}\)

Base-case n=1: 3

^{1}-1=2*1, true for n=1

S(p)=3

^{p}-1=2k

Assuming it is true for n=p, then it is true for n=p+1.

S(p+1)= 3

^{p+1}-1=2r, \(\displaystyle r\in\mathbb{Z}\)

S(p)=3(3

^{p}-1)=3*2k

S(p)=3

^{p+1}-3=6k

S(p)=3

^{p+1}-1=6k+2

S(p)=3

^{p+1}-1=2(3k+1)

(3k+1)=r and hence:

S(p)=3

^{p+1}-1=2r