Question

${lim}_{n\to \infty }\frac{1}{n^3}\left[(1^2+2^2+\dots +n^2)\right]=?$

• A. $1/3$
• B. $1/16$
• C. $1/12$
• D. $1$

Answer 1

The correct option is A $1/3$

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

$={lim}_{x\to \infty }\frac{n(n+1)(2n+1)}{6}$

$=\frac{1}{6}{lim}_{x\to \infty }\left(\frac{n}{n}\right)\left(\frac{n+1}{n}\right)\left(\frac{2n+1}{n}\right)$

$=\frac{1}{6}{lim}_{x\to \infty }1.\left(\frac{n+1}{n}\right)\left(\frac{2n+1}{n}\right)$

$=\frac{1}{3}$

Hence, the answer is $\frac{1}{3}.$