Question

$\sum^n_{r = 1} (2r - 1) = x $ then $\lim_{n \to \infty} \left[ \frac{1^3}{x^2} + \frac{2^3}{x^2} + \frac{3^3}{x^2} + ...... + \frac{n^3}{x^2} \right]= $

• A. 1
• B. $\frac{1}{2}$
• C. 4
• D. $\frac{1}{4}$

Answer 1

Answer: (D)

$x=1+3+5+\dots\dots\dots\dots\dots\dots.. +\left(2n-1\right)$
$x=n^{2}$
$x^{2}=n^{4}$
$\therefore \, lim_{n\rightarrow\infty} \left[\frac{1^{3}+2^{3}+. +n^{3}}{x^{2}}\right]$
$\therefore \, lim_{n\rightarrow\infty} \left[\frac{n^{2}\left(n+1\right)^{2}}{\frac{4}{n^{4}}}\right]$
$lim_{n\rightarrow\infty} \frac{n^{2}.n^{2}\left(1+\frac{1}{n}\right)^{2}}{4n^{4}}\quad\therefore \, lim_{n\rightarrow\infty} \frac{\left(1+\frac{1}{n}\right)^{2}}{4}=\frac{1}{4}$