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I'm trying to separate the sum $$\sum\frac{6^j}{(3^{j+1}-2^{j+1})(3^j-2^j)}$$ into two parts so that I can use the telescoping property. I thought I could use partial fraction decomposition and WolframAlpha supports this idea

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Notice that's exactly the form I need. However I can't figure out how they did it. Once I get to $$6^j = 3^jA-2^jA+3^{j+1}B-2^{j+1}B$$ I'm really not sure how to continue and each way I try gets me something different (and less useful) than what WA gave. How did WA get that result?

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write your numerator as $$3^{j+1}\cdot 2^j-2^{2j+1}-2^{j+1}\cdot 3^j+2^{2j+1}$$ in German "Nulladdition" i don't know the english word

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  • $\begingroup$ Interesting. That definitely works. Thanks! $\endgroup$ – Dylan Apr 11 '17 at 16:50
  • $\begingroup$ As far as I know, we don't have a dedicated word for that idea. "Adding zero" or "add zero" or "add(ing) zero in a clever way" are commonly used. $\endgroup$ – Robert Wolfe Apr 11 '17 at 17:41
  • $\begingroup$ i think adding Zero describes this very good, thank you! $\endgroup$ – Dr. Sonnhard Graubner Apr 11 '17 at 17:42
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HINT:

Let $3^j=a,2^j=b$

$$\frac{6^j}{(3^{j+1}-2^{j+1})(3^j-2^j)}=\dfrac{ab}{(3a-2b)(a-b)}$$

Now $(3a-2b)-2(a-b)=a$ and $(3a-2b)-3(a-b)=b$

Replace the numerator $a$ or $b$ but not both with these values.

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