$6^{(n+2)} + 7^{(2n+1)}$ is divisible by $43$ for $n \ge 1$ Use mathematical induction to prove that 6(n+2) + 7(2n+1) is divisible by 43 for n >= 1.
So start with n = 1:
6(1+2) + 7(2(1)+1) = 63 + 73 = 559 -> 559/43 = 13. So n=1 is divisible
Let P(k): 6(k+2)+7(2k+1) , where k>=1
Show that P(k+1): 6((k+1)+2) + 7(2(k+1)+1) is true
= 6(k+1+2) + 7(2k+2+1)
 = 6(k+3) + 7(2k+3)
I'm unsure where to go from here, I've tried several directions after this but have got nowhere. I don't know how to get the 43 or 559 out the front.
Any help would be great

The sequence $A_n = 6^{n+2} + 7^{2n+1}$ satisfies a two term linear recurrence relation with integer coefficients.  Specifically it satisfies $A_n = 55A_{n-1} - 294A_{n-2}$ but we don't actually care what the relation is.  If $43$ divides $A_n$ for two consecutive $n$ then by induction it must divide every term after that.  So it is enough to check this for say $n = 0$ and $n=1$.
