Question

$\displaystyle\lim_{n\to\infty} \frac{1^{2}+2^{2} +...+n^{2}}{4n^{3}+6n^{2}-5n+1}=$

• A. $\frac{1}{6}$
• B. $\frac{1}{12}$
• C. $\frac{1}{18}$
• D. $\frac{1}{4}$

Answer 1

The Correct Option is B

$\lim_{n\to\infty} \frac{1^{2}+2^{2} +...+n^{2}}{4n^{3}+6n^{2}-5n+1}$
$= \lim _{n\to \infty } \frac{\frac{n\left(+1\right)\left(2n+1\right)}{6}}{4n^{3}+6n^{2}-5n+1} $
$= \frac{1}{6}\lim _{n\to \infty } \left[\frac{2n^{3}+3n^{2}+n}{4n^{3}+6n^{2}-5n+1}\right] $
$=\frac{1}{6}\lim _{n\to \infty } \left[\frac{2+\frac{3}{n}+\frac{1}{n^{2}}}{4+\frac{6}{n}-\frac{5}{n^{2}}+\frac{1}{n^{3}}}\right]$
$ = \frac{1}{6}\left[\frac{2+0+0}{4+0-0+0} =\frac{1}{6} \times\frac{2}{4}=\frac{1}{12}\right]$