Show that 2^(3n)-7n-1 is divisible by 49. Hence show that 2^(3n+3)-7n-8 is divisible by 49?

　
I'm not sure how binomial expansion would help.
Binomial expansion of 2^(3n) is
(1+1)^(3n) = C(3n,0) + C(3n,1) + C(3n,2) + ... + C(3n,3n)

Anyway, we can prove using mathematical induction.

Base case: n = 0
2^(3n) − 7n − 1 = 1 − 0 − 1 = 0 ----> divisible by 49

Inductive Step:
Assume statement is true for some n, then show it is also true for n+1

Since statement is true for n, then: 2^(3n) − 7n − 1 = 49k

For n+1 we get:
2^(3(n+1)) − 7(n+1) − 1
= 2^3 * 2^(3n) − 7n − 7 − 1
= 8*2^(3n) − 56n + 49n − 8
= 8 (2^(3n) − 7n − 1) + 49n
= 8 (49k) + 49n
= 49 (8k+n) ----> divisible by 49

So when statement is true for n, it is also true for n+1

Therefore, by mathematical induction, statement is true for all integers n ≥ 0


Since P(n) = 2^(3n)−7n−1 is divisible by 49, then so is
P(n+1) = 2^(3(n+1)) − 7(n−1) − 1 = 2^(3n+3) − 7n − 8

However, note that P(n+1) was shown to be divisible by 49 in the inductive step of proof above.