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Came across this Devil while preparing for JEE Advanced.
Question: If $$K=\sum_{n=1}^{\infty}\frac{6^n}{(3^n-2^n)(3^{n+1}-2^{n+1})}$$ Then the last digit of $(K+6)^{(K+6)!}$ is?

What i tried to do was to separate $6^n$ as $3^n$$2^n$ and tried to proceed further, but to be honest, I'm getting nowhere around the answer which according to my textbook is 8.

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    $\begingroup$ The last digit of that expression is clearly a $6$. It is followed by a closing parenthesis and an exclamation mark. $\endgroup$ Apr 11, 2017 at 11:33

2 Answers 2

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Hint. One may observe that $$ \frac{6^n}{(3^n-2^n)(3^{n+1}-2^{n+1})}=\frac{2^n}{3^n-2^n}-\frac{2^{n+1}}{3^{n+1}-2^{n+1}} $$ giving a telescoping sum, $N\ge1$, $$ \sum_{n=1}^N\frac{6^n}{(3^n-2^n)(3^{n+1}-2^{n+1})}=2-\frac{2^{N+1}}{3^{N+1}-2^{N+1}}, $$ thus we just have $$ \color{red}{K=2.} $$

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    $\begingroup$ How did you figured that out!? That first step I mean. Is there any method to do that or it's just observation? $\endgroup$ Apr 11, 2017 at 6:16
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    $\begingroup$ @UddeshyaSingh Your observation $6^n=3^n\times2^n$ and seeing the denominator $(3^n-2^n)(3^{n+1}-2^{n+1})$, something clicked. $\endgroup$ Apr 11, 2017 at 6:20
  • $\begingroup$ Now the question is : how does the book gives $8$ for the last digit ? $\endgroup$ Apr 11, 2017 at 6:22
  • $\begingroup$ @ClaudeLeibovici The last digit is not $8$, it is $6$... $\endgroup$ Apr 11, 2017 at 6:24
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    $\begingroup$ I know and this is why I asked ! So, one more typo in a textbook ! $\endgroup$ Apr 11, 2017 at 6:26
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If you really need to see how it telescopes you can do it this way. Take $2^n,2^{n+1} $ common from two brackets now we simplidying we get $$\frac {1}{2}\frac {3^n}{2^n((\frac {3}{2})^n-1)(\frac {3}{2}(\frac{3}{2})^n-1)} $$ now let $(3/2)^n=a $ we see that $\frac {a}{2}=3\frac {a}{2}-1-(a-1) $ thus the series is $\frac {1}{(\frac {3}{2})^n-1}-\frac{1}{\frac {3}{2}(\frac {3}{2})^n-1} $ thus the sum is $2$

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