Q116 / 160 · TS EAMCET 2020 (14 Sep Shift 1) · Mathematics
$\lim _{n \rightarrow \infty}\left[\frac{n+3}{n^2+1^2}+\frac{n+6}{n^2+2^2}+\frac{n+9}{n^2+3^2}+\ldots+\frac{2}{n}\right]=$
- Correct answerA$\frac{\pi}{4}+\frac{3}{2} \ln 2$
- B$\frac{\pi}{2}+\frac{3}{4} \ln 2$
- C$\frac{\pi}{4}-\frac{3}{2} \ln 2$
- D$\frac{\pi}{4}+\frac{1}{2} \ln 2$
Solution
We have,
$\begin{gathered}\lim _{n \rightarrow \infty}\left[\frac{n+3}{n^2+1^2}+\frac{n+6}{n^2+2^2}+\frac{n+9}{n^2+3^2}+\ldots \frac{2}{n}\right] \\ \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n\left(\frac{1+\frac{3 r}{n}}{1+\left(\frac{r}{n}\right)^2}\right)=\int_0^1\left(\frac{1+3 x}{1+x^2}\right) d x \\ =\int_0^1 1\left(\frac{1}{1+x^2}+\frac{3 x}{1+x^2}\right) d x \\ =\left[\tan ^{-1} x+\frac{3}{2} \log \left(1+x^2\right)\right]_0^1 \\ =\tan ^{-1} 1+\frac{3}{2} \log 2=\frac{\pi}{4}+\frac{3}{2} \log 2\end{gathered}$
Hence, the correct answer is option: 'A'.