JEE Main 2022 — Mathematics Algebra

Given,

$$\sum _{k=1}^{10}\frac{k}{{k}^{4}+{k}^{2}+1}=\frac{m}{n}$$k=1∑10​k4+k2+1k​=nm​

$$\Rightarrow \frac{1}{2}\sum _{k=1}^{10}\frac{({k}^{2}+k+1)-({k}^{2}-k+1)}{({k}^{2}+k+1)({k}^{2}-k+1)}=\frac{m}{n}$$⇒21​k=1∑10​(k2+k+1)(k2−k+1)(k2+k+1)−(k2−k+1)​=nm​

$${\text{by using the formula }{x}^{4}+{x}^{2}+1=({x}^{2}+x+1)({x}^{2}-x+1)}$$by using the formula x4+x2+1=(x2+x+1)(x2−x+1)

$$\Rightarrow \frac{1}{2}(\sum _{k=1}^{10}(\frac{1}{({k}^{2}-k+1)}-\frac{1}{{k}^{2}+k+1}))=\frac{m}{n}$$⇒21​(k=1∑10​((k2−k+1)1​−k2+k+11​))=nm​

$$\Rightarrow \frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-.........-\frac{1}{111})=\frac{m}{n}$$⇒21​(1−31​+31​−71​+71​−.........−1111​)=nm​

$$\Rightarrow \frac{1}{2}(1-\frac{1}{111})=\frac{m}{n}$$⇒21​(1−1111​)=nm​

$$\Rightarrow \frac{55}{111}=\frac{m}{n}$$⇒11155​=nm​

So, $m+n=166$m+n=166