Question

$\underset{n\to \infty }{lim}\frac{1^2+2^2+3^2+\dots .+n^2}{2n^3+3n^2+4n+1}=$

• A. $\frac{1}{3}$
• B. $\frac{1}{6}$
• C. $\frac{1}{10}$
• D. $\frac{1}{12}$

Answer 1

The correct option is B $\frac{1}{6}$

Using, the expression for the sum of the squares of the first $n$ consecutive integers,
The given expression simplifies to,
$limn\to \infty \frac{\left(n\right)(n+1)(2n+1)}{6(2n^3+3n^2+4n+1)}$
Dividing the numerator and the denominator by $n^3$
We get the result as $\frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6(2+\frac{3}{n}+\frac{4}{n^2}+\frac{1}{n^3})}$
$=\frac{2}{6\times 2}=\frac{1}{6}$
Hence, option 'B' is correct.