Q117 / 180 · KCET 2019 · Mathematics
\( \sum_{r=1}^{n}(2 r-1)=x \) then
\( \lim _{n \rightarrow \infty}\left[\frac{1^{3}}{x^{2}}+\frac{2^{3}}{x^{2}}+\frac{3^{3}}{\chi^{2}}+\ldots+\frac{n^{3}}{\chi^{2}}\right]= \)
- Correct answerA\( \frac{1}{4} \)
- B\( 04 \)
- C\( \frac{1}{2} \)
- D\( 11 \)
Solution
\( (\mathrm{A}) \)
\( x=1+3+5+\ldots \ldots+(2 n-1) \)
\( x=n^{2} \)
\( x^{2}=n^{4} \)
\( \therefore \lim _{n \rightarrow \infty}\left[\frac{1^{3}+2^{3}+\ldots+n^{3}}{x^{2}}\right] \)
\( \therefore \lim _{n \rightarrow \infty}\left[\frac{\frac{n^{2}(n+1)^{2}}{4}}{n^{4}}\right] \)
\( \therefore \lim _{n \rightarrow \infty}\left[\frac{n^{2} \cdot n^{2}\left(1+\frac{1}{n}\right)^{2}}{4 n^{4}}\right] \)
\( \therefore \lim _{n \rightarrow \infty}\left[\frac{\left(1+\frac{1}{n}\right)^{2}}{4}\right]=\frac{1}{4} \)
Hence, the correct answer is option: 'A'.