9
$\begingroup$

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.

$\endgroup$
1
  • 5
    $\begingroup$ 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

13
$\begingroup$

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 $$ \color{red}{K=2.} $$

$\endgroup$
5
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 2
    $\begingroup$ I know and this is why I asked ! So, one more typo in a textbook ! $\endgroup$ Apr 11, 2017 at 6:26
7
$\begingroup$

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 $2$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.