Question

$\underset{n\to \infty }{lim}\frac{1.n^2+2.(n-1{)}^2+3.(n-2{)}^2+.....n{.1}^2}{1^3+2^3+....n^3}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Given expression is

$\frac{1.n^2+2.(n-1{)}^2+3.(n-2{)}^2+.....n{.1}^2}{1^3+2^3+....n^3}$

Here, we can generalize numerator and denominator as

$\frac{{\sum }_{r=1}^{r=n}[r.(n-r+1{)}^2]}{{\sum }_{r=1}^n{r}^3}$

=$\frac{{\sum }_{r=1}^{r=n}r.(n-r+1{)}^2}{{\sum }_{r=1}^n{r}^3}$

= $\frac{{\sum }_{r=1}^n[r.(n+1{)}^2+r^2-2r(n+1)]}{{\sum }_{r=1}^n{r}^3}$

= $\frac{{\sum }_{r=1}^n[r.(n+1{)}^{2}r+r^3-2r^2(n+1)]}{{\sum }_{r=1}^n{r}^3}$

= $\frac{(n+1{)}^2{\sum }_{r=1}^{n}r+{\sum }_{r=1}^n{r}^3-2(n+1\left){\sum }_{r=1}^n\right(r^2)}{{\sum }_{r=1}^n{r}^3}$

= $\frac{(n+1{)}^2.\frac{n(n+1)}{2}+\frac{n^2(n+1{)}^2}{4}-2(n+1).\frac{n(n+1)(2n+1)}{6}}{\frac{n^2(n+1{)}^2}{4}}$

Dividing numerator and denominator by $\frac{n(n+1{)}^2}{2}$
$=\frac{\frac{(n+1)}{4}+\frac{n}{8}-\frac{(2n+1)}{6}}{\frac{n}{8}}$
$\underset{n\to \infty }{lim}\frac{\frac{1}{4}+\frac{1}{8}-\frac{1}{3}-\frac{1}{6n}+\frac{1}{4n}}{\frac{1}{8}}$
$=\frac{\frac{1}{24}}{\frac{1}{8}}$
= 1/3