Question

$L{t}_{n\to \infty }\frac{1.n^2+2(n-1{)}^2+3(n.-2{)}^2+\dots +n{.1}^2}{1^3+2^3+..+n^3}$ equals to:

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

Answer 1

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

$L{t}_{n\to \infty }\frac{1.n^2+2(n-1{)}^2+3(n.-2{)}^2++n{.1}^2}{1^3+2^3+..+n^3}$

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

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

$=\underset{{}_{n\to \infty }}{lim}\frac{\frac{n^2{(n+1)}^2}{4}+{(n+1)}^2\frac{n(n+1)}{2}-(2n+1)\frac{n(n+1)(2n+1)}{6}}{\frac{n^2{(n+1)}^2}{4}}$

$=1+2-\frac{8}{3}=\frac{1}{3}$