Question

$\underset{n\to \infty }{\text{lim}}\frac{1\cdot n^2+2{(n-1)}^2+3{(n-2)}^2+\cdots +n\cdot 1^2}{1^3+2^3+\cdots +n^3}$ is equal to

• A. $\frac{8}{3}$
• B. $\frac{4}{3}$
• C. $\frac{2}{3}$
• D. $\frac{1}{3}$

Answer 1

The correct option is D $\frac{1}{3}$

Given expansion

$=\frac{\sum _{r=1}^n[r\cdot {(n-r+1)}^2]}{\sum _{r=1}^n{r}^3}=\underset{n\to \infty }{\text{lim}}\frac{\sum _{r=1}^n{(n+1)}^{2}r-2(n+1)r^2+r^3}{\frac{n^2{(n+1)}^2}{4}}$

$=\underset{n\to \infty }{\text{lim}}\frac{{(n+1)}^2\frac{n(n+1)}{2}-2(n+1)\frac{n(n+1)(2n+1)}{6}+\frac{n^2{(n+1)}^2}{4}}{\frac{n^2{(n+1)}^2}{4}}$
$\because$ Degree of numerator $=$Degree of denominator
Given limit $=\frac{\frac{1}{2}-\frac{2}{3}+\frac{1}{4}}{\frac{1}{4}}$
$=2-\frac{8}{3}+1=\frac{9-8}{3}=\frac{1}{3}$