Question

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

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

Answer 1

The correct option is B $-\frac{1}{3}$

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

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

${lim}_{n\to \infty }\frac{1}{1-n^3}{\sum }_{t=1}^n{r}^2$
$={lim}_{n\to \infty }\frac{1}{n^3(\frac{1}{n^3}-1)}.\frac{n(n+1)(2n+1)}{6}$
$=-\frac{1}{3}$