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
Sum of series $\frac85+\frac{16}{65}+\cdots \cdots +\cdots +\frac{128}{2^{18}+1}$
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sequences-and-series
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3these are very few terms, you can easily sum them using a simple calculator. – 2017-01-04
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0Hint: $2^{5+2}=128$. – 2017-01-04
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0I think he means $\sum_k a_k$ – 2017-01-04
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0[There you go](http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bn%3D1%7D%5E%7B5%7D%5Cfrac%7B2%5E%7Bn%2B2%7D%7D%7B4%5E%7B2n-1%7D%2B1%7D). Anything else? – 2017-01-04
1 Answers
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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} $