Question

If ${\mathrm{S}}_{\mathrm{n}}=\frac{1}{1^3}+\frac{1+2}{1^3+2^3}+\frac{1+2+3}{1^3+2^3+3^3}+\dots ..+\frac{1+2+\dots \dots +\mathrm{n}}{1^3+2^3+\dots \dots +{\mathrm{n}}^3};n=1,2,3..$ Then $S_n$ is not greater than:

• A. $\frac{1}{2}$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is C

$2$

Explanation for the correct option.

Step 1: Find the rth term

Given, ${\mathrm{S}}_{\mathrm{n}}=\frac{1}{1^3}+\frac{1+2}{1^3+2^3}+\frac{1+2+3}{1^3+2^3+3^3}+\dots ..+\frac{1+2+\dots \dots +\mathrm{n}}{1^3+2^3+\dots \dots +{\mathrm{n}}^3};n=1,2,3..$

$\begin{array}{rcl}{S}_n& =& \frac{{\displaystyle \underset{k=1}{\overset{r}{\sum k}}}}{\underset{k=1}{\overset{r}{\sum k^3}}}\\ & =& \frac{{\displaystyle \frac{r(r+1)}{2}}}{{\left[\frac{r(r+1)}{2}\right]}^2}\\ & =& \frac{2}{r(r+1)}\\ & =& \frac{2}{r}-\frac{2}{(r+1)}\end{array}$

Step 2: Find the sum

$\begin{array}{rcl}{S}_n& =& \underset{r=1}{\overset{n}{\sum \frac{2}{r}-\left(\frac{2}{r+1}\right)}}\\ & =& \frac{2}{1}-\frac{2}{1}+\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-,,,\frac{2}{n}-\frac{2}{n+1}\\ & =& 2-\frac{2}{n+1}<2\end{array}$

Hence, option C is correct.