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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$$2^n$ and tried to proceed further, but to be honest, I'm getting nowhere around the answer which according to my textbook is 8.
Hint. One may observe that $$ \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.} $$
If you really need to see how it telescopes you can do it this way. Take $2^n,2^{n+1} $ common from two brackets now we simplidying we get $$\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 $2$
