Using the principle of mathematical induction, prove that `(2^(3n)-1)` is divisible by `7` for all `n in Ndot`

Explanation

Let P(n) : `2^(3n)-1` divisible by 7
Step I We observe that P(1) is true.
`P(1):2^(3xx1)-1=2^(3)-1=8-1=7`
It is clear that P(1) is true.
Stem II Now, assume that P(n) is true for n=k,
`P(k):2^(3k)-1` is divisible by 7.
`rArr2^(3k)-1=7q`
Stem III Now, to prove P(k+1)is true. `P(k+1):2^(3(k+1))-1`
`=2^(3k).2^(3)-1`
`=2^(3k)(7+1)-1`
`=7*2^(3k)+2^(3k)-1`
`7*2^(3k)++7q` [from stepII]
`=7(2^(3k)+q)`