Originally Posted by

**MrT83** Question:

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.