lim _{n rightarrow infty} frac{1}{n}left{sin | vedclass

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(D) The given expression is $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{3n} \sin ^5\left(\frac{r \pi}{6 n}\right)$.By the definition of defi

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$\frac{16}{5 \pi}$

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(D) The given expression is $\lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^{3n} \sin ^5\left(\frac{r \pi}{6 n}ight)$.By the definition of definite integral as a limit of a sum,$\lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^{kn} f\left(\frac{r}{n}ight) = \int_0^k f(x) dx$.Here,$f(x) = \sin^5\left(\frac{\pi}{6} xight)$ and $k=3$.So,the integral is $\int_0^3 \sin ^5\left(\frac{\pi}{6} xight) dx$.Let $t = \frac{\pi}{6} x$,then $dt = \frac{\pi}{6} dx$,which implies $dx = \frac{6}{\pi} dt$.When $x=0, t=0$ and when $x=3, t=\frac{\pi}{2}$.The integral becomes $\frac{6}{\pi} \int_0^{\pi/2} \sin^5(t) dt$.Using Wallis' formula,$\int_0^{\pi/2} \sin^n(x) dx = \frac{(n-1)!!}{n!!} 	imes \frac{\pi}{2}$ (if $n$ is even) or $\frac{(n-1)!!}{n!!}$ (if $n$ is odd).For $n=5$,$\int_0^{\pi/2} \sin^5(t) dt = \frac{4 	imes 2}{5 	imes 3 	imes 1} = \frac{8}{15}$.Therefore,the value is $\frac{6}{\pi} 	imes \frac{8}{15} = \frac{2 	imes 8}{5 \pi} = \frac{16}{5 \pi}$.

$\lim _{n \rightarrow \infty} \frac{1}{n}\left\{\sin ^5\left(\frac{\pi}{6 n}\right)+\sin ^5\left(\frac{2 \pi}{6 n}\right)+\sin ^5\left(\frac{3 \pi}{6 n}\right)+\ldots+\sin ^5\left(\frac{\pi}{2}\right)\right\} = $

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