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$\sum\limits_{n=1}^{\infty}\frac{6^n}{(3^{n+1}-2^{n+1})(3^n-2^n)}$ I saw this and immediately thought it was a telescoping series, so I tried to partial fraction it. $$\frac{6^n}{(3^{n+1}-2^{n+1})(3^n-2^n)}=\frac{A}{3^{n+1}-2^{n+1}}+\frac{B}{3^n-2^n}$$ \begin{align} 6^n &= A(3^n-2^n)+B(3^{n+1}-2^{n+1})\\ 6^n &=3^n(A+3B)+2^n(-A-2B) \end{align} This is where I got stuck, how can the sum of these 2 be $6^n$ so I figured one way it could be true was if $$A+3B=\frac{2^n}{2}$$ $$-A-2B=\frac{3^n}{2}$$ Because then I will just be doing $\frac{6^n}{2}+\frac{6^n}{2}=6^n$ Solving this system gave me \begin{align} A&=-\frac{3^{n+1}+2^{n+1}}{2}\\ B&=\frac{3^n+2^n}{2} \end{align} As terrible as it looks, I got that it did telescope to $\frac{5}{2}$ however it actually came out to $2$ with the partial fraction of $$\frac{2^n}{3^n-2^n}-\frac{2^{n+1}}{3^{n+1}-2^{n+1}}$$ So 2 questions, Why am I wrong? and What would be the correct approach to get this answer?

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  • $\begingroup$ The only way I can see you getting $\frac52$ is by mistakenly substituting $n=1$ into $B$ rather than into $\frac{2^n}{3^n-2^n}$. $\endgroup$ – Mike Mar 19 '17 at 21:13
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Is there a transcription problem? A check on your work shows $$ \frac{6^n}{\left(3^{n+1}-2^{n+1}\right) \left(3^n-2^n\right)} = \frac{2^n}{3^n-2^n}-\frac{2^{n+1}}{3^{n+1}-2^{n+1}}, $$ and $$ \sum _{n=1}^{\infty } \frac{6^n}{\left(3^{n+1}-2^{n+1}\right) \left(3^n-2^n\right)} = 2. $$

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    $\begingroup$ Yes I know, I'm wondering as to why I am wrong, I wrote that partial fraction at the end of the question $\endgroup$ – Teh Rod Mar 19 '17 at 19:56
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Hint: write your term in th form $$\frac{2^n}{3^n-2^n}-\frac{2^{n+1}}{3^{n+1}-2^{n+1}}$$

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  • $\begingroup$ How did you get that? I don't know how to go straight to that $\endgroup$ – Teh Rod Mar 19 '17 at 19:56

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