Prove by induction that 4^n + 6n + 8 is divisible by 9 for all positive integers n.
My attempted solution:
for n = 1
4^n + 6n + 8 = 4 + 6 + 8 = 18 = 9.2
so is divisible by 9 for n = 1.
Suppose true for n = k, k is a natural number
then 4^k + 6k + 8 = 9p, where p is an integer.
Consider when n = k + 1
then we have
4^(k + 1) + 6(k + 1) + 8
= 4.4^k + 6k + 6 + 8
= 3.4^k + 4^k + 6k + 8 + 6
= 3.4^k + 6 + 9p
Which can easily be shown to be divisible by 3 but not so easily for 9.
Any help/advice would be greatly appreciated.
Apologies for not using LATEX - it's been a few years since I used it and I couldn't make it work properly.