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?

2. Originally Posted by zhupolongjoe
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?
Hi Polong,

You are using induction. Let's start from where you were.

$f_{k+1} =8 \times (-5 \times 3^k+5^{k-1}+2^{k+3})-15 \times (-5 \times 3^{k-1}+5^{k-2}+2^{k+2})+6\times 2^{k+1}$
$=-\frac{40}{3}\times 3^{k+1}+\frac{8}{5} \times 5^k+4 \times 2^{k+4}+\frac{25}{3}\times 3^{k+1}-\frac{3}{5} \times 5^k-\frac{15}{4}\times 2^{k+4}+\frac{3}{4}\times 2^{k+4}$
$=\left(-\frac{40}{3}+\frac{25}{3}\right)\times 3^{k+1}+\left(\frac{8}{5}-\frac{3}{5}\right) \times 5^k+\left(4-\frac{15}{4}+\frac{3}{4}\right) \times 2^{k+4}$
$=-5 \times 3^{k+1}+5^k+2^{k+4}$

Good luck!