Correct answer is : 32
\(\sum\limits_{r=1}^{\infty} \frac{n}{\sqrt{n^{4}+r^{4}}}-\frac{2 n r^{2}}{\left(n^{2}+r^{2}\right) \sqrt{n^{4}+r^{4}}}\)
\(\sum\limits_{r=1}^{\infty} \frac{\frac{1}{n}}{\sqrt{1+\left(\frac{r}{n}\right)^{4}}}-\frac{2\left(\frac{1}{n}\right)\left(\frac{r}{n}\right)^{2}}{\left(1+\left(\frac{r}{n}\right)^{2}\right) \sqrt{1+\left(\frac{r}{n}\right)^{4}}}\)
\(\Rightarrow \int\limits_{0}^{1} \frac{d x}{\sqrt{1+x^{4}}}-\frac{2 x^{2} d x}{\left(1+x^{2}\right) \sqrt{1+x^{4}}}\)
\(\Rightarrow \int\limits_{0}^{1} \frac{1-x^{2}}{\left(1+x^{2}\right) \sqrt{1+x^{4}}} d x\)
\(\Rightarrow \int\limits_{0}^{1} \frac{\frac{1}{x^{2}}-1}{\left(x+\frac{1}{x}\right) \sqrt{x^{2}+\frac{1}{x^{2}}}} d x\)
\(\Rightarrow-\int\limits_{0}^{1} \frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right) \sqrt{\left(x+\frac{1}{x}\right)^{2}-2}} d x\)
\(\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{t} \Rightarrow 1-\frac{1}{\mathrm{x}^{2}} \mathrm{dx}=\mathrm{dt}\)
\( \Rightarrow-\int\limits_{\infty}^{2} \frac{\mathrm{dt}}{\mathrm{t} \sqrt{\mathrm{t}^{2}-2}}\)
\( \Rightarrow-\int\limits_{\infty}^{2} \frac{\mathrm{tdt}}{\mathrm{t}^{2} \sqrt{\mathrm{t}^{2}-2}}\)
\(\text{take} \ \mathrm{t}^{2}-2=\alpha^{2}\)
\(\mathrm{t\ dt}=\alpha \ \mathrm{d} \alpha\)
\(\Rightarrow-\int\limits_{\infty}^{\sqrt{2}} \frac{\alpha \mathrm{d} \alpha}{\left(\alpha^{2}+2\right) \alpha}\)
\(\Rightarrow-\int\limits_{\infty}^{\sqrt{2}} \frac{\mathrm{d} \alpha}{\alpha^{2}+2}\)
\(\left.\Rightarrow \frac{-1}{\sqrt{2}} \tan ^{-1} \frac{\alpha}{\sqrt{2}}\right]_{\infty}^{\sqrt{2}}\)
\(\Rightarrow \frac{-1}{\sqrt{2}}\left\{\tan ^{-1} 1\right\}+\frac{1}{\sqrt{2}} \tan ^{-1} \infty\)
\(\Rightarrow \frac{1}{\sqrt{2}}\left\{\frac{\pi}{2}-\frac{\pi}{4}\right\}\)
\(\Rightarrow \frac{\pi}{4 \sqrt{2}}=\frac{\pi}{\mathrm{K}}\)
\(\text{So} \ \mathrm{K}=4 \sqrt{2}\)
\(\mathrm{K}^{2}=32\)
