# Infinite sum proof by induction

1. Nov 12, 2007

### Doom of Doom

So I am trying to solve a problem.

Evaluate $$\sum_{k=1}^{\infty}\frac{6^{k}}{(3^{k}-2^{k})(3^{k+1}-2^{k+1})}$$.

Essentially, I've boiled it down to this, but I can't quite prove it:
$$\sum_{k=1}^{n}\frac{6^{k}}{(3^{k}-2^{k})(3^{k+1}-2^{k+1})}=\frac{6(3^{n}-2^{n})}{(3^{n+1}-2^{n+1})}$$
and the limit of this as n approaches infinity is 2.

I need to be able to prove that $$3^{n+1}-2^{n+1}+6^{n-1}=(3^{n}-2^{n})^{2}$$ in order for my induction hypothesis to work, and I'm having trouble for some reason. Help?

2. Nov 12, 2007

### d_leet

Perhaps beacuse it isn't true, consider n=3 then

3^(3+1)-2^(3+1)+6^(3-1)=81-16+36=101
whereas
(3^3-2^3)^2=(27-8)^2=(19)^2=361

3. Nov 12, 2007

### Doom of Doom

Wow, I'm sorry. Perhaps that's why I was unable to do it. My notes are all over the place and I guess I kind of lost track of a lot of numbers somewhere.

I meant to have $$(3^{n-1}-2^{n-1})(3^{n+1}-2^{n+1})+6^{n-1}=(3^{n}-2^{n})^{2}$$

Now I'm able to prove it! Thanks anyway.