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KCET · Maths · Limits

\( \sum_{r=1}^{n}(2 r-1)=x \) then
\( \lim _{n \rightarrow \infty}\left[\frac{1^{3}}{x^{2}}+\frac{2^{3}}{x^{2}}+\frac{3^{3}}{\chi^{2}}+\ldots+\frac{n^{3}}{\chi^{2}}\right]= \)

  1. A \( \frac{1}{4} \)
  2. B \( 04 \)
  3. C \( \frac{1}{2} \)
  4. D \( 11 \)
Verified Solution

Answer & Solution

Correct Answer

(A) \( \frac{1}{4} \)

Step-by-step Solution

Detailed explanation

\( (\mathrm{A}) \)
\( x=1+3+5+\ldots \ldots+(2 n-1) \)
\( x=n^{2} \)
\( x^{2}=n^{4} \)
\( \therefore \lim _{n \rightarrow \infty}\left[\frac{1^{3}+2^{3}+\ldots+n^{3}}{x^{2}}\right] \)
\( \therefore \lim _{n \rightarrow \infty}\left[\frac{\frac{n^{2}(n+1)^{2}}{4}}{n^{4}}\right] \)
\( \therefore \lim _{n \rightarrow \infty}\left[\frac{n^{2} \cdot n^{2}\left(1+\frac{1}{n}\right)^{2}}{4 n^{4}}\right] \)
\( \therefore \lim _{n \rightarrow \infty}\left[\frac{\left(1+\frac{1}{n}\right)^{2}}{4}\right]=\frac{1}{4} \)