Question

${\sum }_{r=1}^n(2r-1)=x$ 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)

$x=1+3+5+\dots \dots \dots \dots \dots \dots ..+\left(2n-1\right)$
$x=n^2$
$x^2=n^4$
$\therefore li{m}_{n\to \infty }\left[\frac{1^3+2^3+.+n^3}{x^2}\right]$
$\therefore li{m}_{n\to \infty }\left[\frac{n^2{\left(n+1\right)}^2}{\frac{4}{n^4}}\right]$
$li{m}_{n\to \infty }\frac{n^2.n^2{\left(1+\frac{1}{n}\right)}^2}{4n^4}\therefore li{m}_{n\to \infty }\frac{{\left(1+\frac{1}{n}\right)}^2}{4}=\frac{1}{4}$