Question

If $S=\frac{8}{5}+\frac{16}{65}+\frac{24}{325}+\dots +\frac{128}{4^9+1}$, then $S=$

• A. $\frac{1024}{545}$
• B. $\frac{1088}{1056}$
• C. $\frac{1088}{545}$
• D. $\frac{1024}{45}$

Answer 1

The correct option is C

$\frac{1088}{545}$

$S={\sum }_{r=1}^{16}\frac{8r}{4r^4+1}$

$={\sum }_{r=1}^{16}\frac{8r}{(2r^2-2r+1)(2r^2+2r+1)}$

$=2{\sum }_{r=1}^{16}\frac{1}{2r^2-2r+1}-\frac{1}{2r^2+2r+1}$

$=2[(1-\frac{1}{5})+(\frac{1}{5}-\frac{1}{13})+(\frac{1}{13}-\frac{1}{25})+\dots +(\frac{1}{481}-\frac{1}{545})]$

$=2(1-\frac{1}{545})=\frac{1088}{545}$