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Q.
The $\operatorname{sum} \displaystyle\sum_{ k =1}^{100} \frac{ k }{ k ^4+ k ^2+1}$ is equal to $\frac{ N }{10101}$ their find the value of $N$.
Sequences and Series
Solution:
$T _{ k }=\frac{ k }{\left( k ^2+1\right)^2- k ^2}=\frac{ k }{\left( k ^2+1+ k \right)\left( k ^2+1- k \right)}$
$=\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(\displaystyle\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)}$
$\text { put } n=100 \text { to get } \frac{5050}{10101} \Rightarrow N=5050$