Question

$\sum _{k=1}^n\frac{k}{k^4+\frac{1}{4}}$ =

• A. $\frac{4n^2-4n+4}{2n^2+2n+1}$
• B. $\frac{3n^2-n+2}{2n^2+2n+1}$
• C. $\frac{2n^2+2n}{2n^2+2n+1}$
• D. $\frac{n^2+5n-2}{2n^2+2n+1}$

Answer 1

The correct option is C $\frac{2n^2+2n}{2n^2+2n+1}$

$S_n={\sum }_{k=1}^n\frac{k}{k^4+\frac{1}{4}}={\sum }_{k=1}^n\frac{4k}{(4k^4+1)}{S}_n={\sum }_{k=1}^n\frac{(2k^2+2k+1)-(2k^2-2k+1)}{(2k^2+2k+1)(2k^2-2k+1)}{S}_n={\sum }_{k=1}^n[\frac{1}{(2k^2-2k+1)}-\frac{1}{(2k^2-2k+1)}]S_n=\{\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+\frac{1}{13}+...+\frac{1}{2n^2-2n+1}-\frac{1}{2n^2+2n+1}\}{S}_n=1-\frac{1}{2n^2+2n+1}=\frac{2n^2+2n+1-1}{2n^2+2n+1}{S}_n=\frac{2n^2+2n}{2n^2+2n+1}$