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sum of series $\displaystyle \frac{8}{5}+\frac{16}{65}+\cdots \cdots +\frac{128}{2^{18}+1}$

I have calculate $\displaystyle a_{n} = \frac{2^{n+2}}{4^{2n-1}+1}$

could some help me with this, thanks

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    $\begingroup$ these are very few terms, you can easily sum them using a simple calculator. $\endgroup$
    – Math-fun
    Commented Jan 4, 2017 at 9:09
  • $\begingroup$ Hint: $2^{5+2}=128$. $\endgroup$ Commented Jan 4, 2017 at 9:11
  • $\begingroup$ I think he means $\sum_k a_k$ $\endgroup$
    – Alex
    Commented Jan 4, 2017 at 9:11

1 Answer 1

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We can write $\displaystyle a_{n} = \frac{2^{n+2}}{4^{2n-1}+1} $ as $\displaystyle T_{n} = \frac{16.2^{n}}{2^{4n}+4} $

$\displaystyle T_{n} = \frac{4}{2^{2n}-2.2^{n}+2}-\frac{4}{2^{2n}+2.2^{n}+2} $

Adding upto n terms we get $\displaystyle S_{n} = 2-\frac{4}{2^{2n}+2.2^{n}+2} $ as it is telescopic

Solving n=5

$\displaystyle S_{5} = 2-\frac{4}{1090}=\frac{1088}{545} $

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  • $\begingroup$ @DXT The answer is correct, if you are satisfied with it, please close this problem $\endgroup$ Commented Jan 28, 2019 at 2:07

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