Q115 / 160 · TS EAMCET 2020 (14 Sep Shift 1) · Mathematics
$\begin{array}{r}\lim _{n \rightarrow \infty}\left[\frac{n^{3 / 2}}{n^{5 / 2}}-\frac{n^{1 / 2}}{n^{3 / 2}}+\frac{n^{3 / 2}}{(n+2)^{5 / 2}}-\frac{n^{1 / 2}}{(n+3)^{3 / 2}}\right. \\ +\frac{n^{3 / 2}}{(n+4)^{5 / 2}}-\frac{n^{1 / 2}}{(n+6)^{3 / 2}}+\ldots .+\frac{n^{3 / 2}}{(n+2(n-1))^{5 / 2}} \\ \left.\quad-\frac{n^{1 / 2}}{(n+3(n-1))^{3 / 2}}\right]=\end{array}$
- A$\frac{-\sqrt{2}}{3}$
- Correct answerB$\frac{-1}{9 \sqrt{3}}$
- C$\frac{\sqrt{2}}{3}$
- D$\frac{1}{9 \sqrt{3}}$
Solution
$\begin{gathered}\lim _{n-\infty}\left[\frac{n^{3 / 2}}{n^{5 / 2}}-\frac{n^{1 / 2}}{n^{3 / 2}}+\frac{n^{3 / 2}}{(n+2)^{5 / 2}}-\frac{n^{1 / 2}}{(n+3)^{3 / 2}}+\ldots\right. \\ \left.\frac{n^{3 / 2}}{(n+2)(n-1))^{5 / 2}}-\frac{n^{1 / 2}}{(n+3(n-1))^{3 / 2}}\right]\end{gathered}$
$\begin{aligned} & =\lim _{n \rightarrow \infty} \frac{1}{n}\left[\sum_{r=0}^{n-1}\left(\frac{1}{\left(1+\frac{2 r}{n}\right)^{5 / 2}}-\frac{1}{\left(1+\frac{3 r}{n}\right)^{3 / 2}}\right)\right] \\ & =\int_0^1\left(\frac{1}{(1+2 n)^{5 / 2}}-\frac{1}{(1+3 n)^{3 / 2}}\right) d x \\ & =\left[\frac{-1}{3} \frac{1}{(1+2 n)^{3 / 2}}+\frac{2}{3(1+3 n)^{1 / 2}}\right]_0^1 \\ & =-\frac{1}{3} \times \frac{1}{3 \sqrt{3}}+\frac{1}{3}+\frac{2}{3 \times 2}-\frac{2}{3} \\ & =-\frac{1}{9 \sqrt{3}}+\frac{1}{3}+\frac{1}{3}-\frac{2}{3}=-\frac{1}{9 \sqrt{3}}\end{aligned}$
Hence, the correct answer is option: 'B'.