Question

If $S_1=\sum n,S_2=\sum n^2$ and $S_3=\sum n^3$, then the value of ${lim}_{n\to \infty }\frac{S_1(1+S_3/8)}{S_2^2}$ is equal to

• A. $\frac{3}{32}$
• B. $\frac{3}{64}$
• C. $\frac{9}{32}$
• D. $\frac{9}{64}$

Answer 1

The correct option is D $\frac{9}{64}$

We have
$S_1=\sum n=\frac{n(n+1)}{2}$
$S_2=\sum n^2=\frac{n(n+1)(2n+1)}{6}$
$S_3=\sum n^3={\left\{\frac{n(n+1)}{2}\right\}}^2$
Therefore, $\underset{n\to \infty }{lim}\frac{S_1(1+\frac{S_3}{8})}{S_2^2}=\underset{n\to \infty }{lim}\frac{\frac{n(n+1)}{2}(1+\frac{n^2{(n+1)}^2}{32})}{\frac{n^2{(n+1)}^2{(2n+1)}^2}{36}}$
$=\underset{n\to \infty }{lim}\frac{18}{32}\times \frac{\left\{32n^2{(n+1)}^2\right\}}{n(n+1){(2n+1)}^2}=\frac{9}{16}\underset{n\to \infty }{lim}\frac{\frac{32}{n^4}+{(1+\frac{1}{n})}^2}{(1+\frac{1}{n}){(2+\frac{1}{n})}^2}$
$=\frac{9}{16}\times \frac{1}{1\times 4}=\frac{9}{64}$