Question

The $\mathrm{sum}\sum _{k=1}^{100}\frac{k}{k^4+k^2+1}$ is equal to $\frac{N}{10101}$ their find the value of $N$.

Answer 1

Answer: 5050

$T_k=\frac{k}{{\left(k^2+1\right)}^2-k^2}=\frac{k}{\left(k^2+1+k\right)\left(k^2+1-k\right)}$
$=\frac{1}{2}\left(\frac{\left(k^2+1+k\right)-\left(k^2+1-k\right)}{\left(k^2+1+k\right)\left(k^2+1-k\right)}\right)=\frac{1}{2}\left(\frac{1}{k^2+1-k}-\frac{1}{k^2+1+k}\right)$
$=\frac{1}{2}\left(\sum _{k=1}^n\frac{1}{k^2+1-k}-\frac{1}{k^2+1+k}\right)$
$=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right)$
$+\frac{1}{2}\left(\frac{1}{3}-\frac{1}{7}\right)$
$+\frac{1}{2}\left(\frac{1}{n^2+1-n}-\frac{1}{n^2+1+n}\right)$
$=\frac{1}{2}\left(1-\frac{1}{n^2+1+n}\right)=\frac{1}{2}\left(\frac{n^2+1+n-1}{n^2+1+n}\right)=\frac{n(n+1)}{2\left(n^2+n+1\right)}$
$\text{ put }n=100\text{ to get }\frac{5050}{10101}\Rightarrow N=5050$