Question

The sum of the series $\frac{8}{5}+\frac{16}{65}+...+\frac{128}{2^{18}+1}$ is

• A. $\frac{540}{1088}$
• B. $\frac{1088}{545}$
• C. $\frac{1001}{500}$
• D. $\frac{1013}{545}$

Answer 1

The correct option is B $\frac{1088}{545}$

Since $S_n={\sum }_{r=1}^n\frac{8r}{4r^4+1}={\sum }_{r=1}^n\frac{8r}{(2r^2-2r-1)(2r^2+2r+1)}$

$=2{\sum }_{r=1}^n\frac{(2r^2+2r+1)-(2r^2-2r+1)}{(2r^2-2r+1)(2r^2+2r+1)}=2{\sum }_{r=1}^n(\frac{1}{2r^2-2r+1}-\frac{1}{2r^2+2r+1})$

$=2\{1-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+.....+\frac{1}{2n^2-2n+1}-\frac{1}{2n^2+2n+1}\}$

$=2\{1-\frac{1}{2n^{+}2n+1}\}=\frac{2(2n^2+2n)}{(2n^2+2n+1)}=\frac{4n^2+2n}{(2n^2+2n+1)}$

$\therefore S_{16}=\frac{4(16{)}^2+4(16)}{2(16{)}^2+2(16)+1}=\frac{1088}{545}$