Define f:N-->N by f(1)=2, f(2)=-8 and for n>=3, f(n)=8f(n-1)-15f(n-2)+6*2^n.

Prove that for all natural numbers n, f(n)=-5*3^n+5^(n-1)+2^(n+3)

I first see that P(1) and P(2) are true to establish the base case.

Next I put f(i)=-5*3^i+5^(i-1)+2^(i+3) for 1<=i<=k.

So f(k+1) is 8f(k)-15f(k-1)+6*2^(k+1)

=8(-5*3^k+5^(k-1)+2^(k-3))-15(-5*3^(k-1)+5^(k-2)+2^(k+2))+6*2^(k+1)

Now is where I am lost? Am I right up to here and where do I go now?