Question

$\underset{n\to \infty }{lim}\frac{1^{2}n+2^2(n-1)+3^2(n-2)+\cdots +n^2.1}{1^3+2^3+3^3+\dots \cdots +n^3}$ is equal to

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{2}$
• D. $\frac{1}{6}$

Answer 1

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

${lim}_{n\to \infty }\frac{1^{2}n+2^2(n-1)+3^2(n-2)+\cdots +n^2.1}{1^3+2^3+3^3+\dots \cdots +n^3}={lim}_{n\to \infty }\frac{{\sum }_{r=1}^n\left(r^2\right)(n-r+1)}{{\sum }_{r=1}^n{r}^3}={lim}_{n\to \infty }\frac{\frac{n^2(n+1)(2n+1)}{6}-{\left(\frac{n(n+1)}{2}\right)}^2+\frac{n(n+1)(2n+1)}{6}}{{\left(\frac{n(n+1)}{2}\right)}^2}={lim}_{n\to \infty }(\frac{2(2n+1)}{3(n+1)}-1+\frac{2(2n+1)}{3n(n+1)})=\frac{4}{3}-1+0=\frac{1}{3}$
Hence, option 'A' is correct.