- #1
Doom of Doom
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So I am trying to solve a problem.
Evaluate [tex]\sum_{k=1}^{\infty}\frac{6^{k}}{(3^{k}-2^{k})(3^{k+1}-2^{k+1})}[/tex].
Essentially, I've boiled it down to this, but I can't quite prove it:
[tex]\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})}[/tex]
and the limit of this as n approaches infinity is 2.
I need to be able to prove that [tex]3^{n+1}-2^{n+1}+6^{n-1}=(3^{n}-2^{n})^{2}[/tex] in order for my induction hypothesis to work, and I'm having trouble for some reason. Help?
Evaluate [tex]\sum_{k=1}^{\infty}\frac{6^{k}}{(3^{k}-2^{k})(3^{k+1}-2^{k+1})}[/tex].
Essentially, I've boiled it down to this, but I can't quite prove it:
[tex]\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})}[/tex]
and the limit of this as n approaches infinity is 2.
I need to be able to prove that [tex]3^{n+1}-2^{n+1}+6^{n-1}=(3^{n}-2^{n})^{2}[/tex] in order for my induction hypothesis to work, and I'm having trouble for some reason. Help?