For positive integers K=1, 2, 3, ……,n. Let $S_k$ denotes the area of $\Delta OA{B}_K$ (where ‘O’ is the origin) such that $\angle AO{B}_K=\frac{K\pi }{2n},OA=1$ and $O{B}_K=K.$ The value of the $\underset{n\to \infty }{\lim }\frac{{\pi }^2}{n^2}.\sum _{K=1}^n\frac{S_K}{10}$ is

Here, $O{B}_K=K$  and $\angle AO{B}_K=\frac{K\pi }{2n}$

$\therefore S_K=\frac{1}{2}.\left(1\right)\left(K\right)\sin \left(\frac{Kn}{2n}\right)[\text{using},\Delta =\frac{1}{2}ab\sin \theta ]$

Then,  $L=\underset{x\to \infty }{\lim }\frac{1}{n^2}\sum _{K=1}^n\frac{K}{2}\sin \left(\frac{K\pi }{2n}\right)$

$$\begin{gathered} =\frac{1}{2}.\underset{n\to \infty }{\lim }\frac{K}{n}.\sin (\frac{\pi }{2}.\frac{K}{n})=\frac{1}{2}.{\int }_0^{1}x.\sin \left(\frac{\pi }{2}x\right)dx \\ =\frac{1}{2}[{\left(\underset{0}{\underbrace{\frac{-2}{\pi }.x.\cos \frac{\pi x}{2}}}\right)}_0^1+\frac{2}{\pi }{\int }_0^1\cos \frac{\pi x}{2}.dx] \\ =\frac{1}{2}[\frac{2}{\pi }.\frac{2}{\pi }.{\left(\sin \frac{\pi x}{2}\right)}_0^1]=\frac{2}{{\pi }^2} \\ \therefore \frac{{\pi }^2}{10}L=\frac{\cancel{{\pi }^2}}{10}.\frac{2}{\cancel{{\pi }^2}} \\ =\frac{1}{5}=0.2 \end{gathered}$$