# Sum of series $\frac85+\frac{16}{65}+\cdots \cdots +\cdots +\frac{128}{2^{18}+1}$

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

• these are very few terms, you can easily sum them using a simple calculator. – Math-fun Jan 4 '17 at 9:09
• Hint: $2^{5+2}=128$. – Olivier Oloa Jan 4 '17 at 9:11
• I think he means $\sum_k a_k$ – Alex Jan 4 '17 at 9:11
• There you go. Anything else? – barak manos Jan 4 '17 at 9:16

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