Question

The sum of $1^{st}$ $n$ terms of the series $\frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{1+2}+\frac{1^{2}+2^{2}+3^{2}}{1+2+3} + ...$

• A. $\ce{\frac{n + 2}{3}}$
• B. $\ce{\frac{n(n + 2)}{3}}$
• C. $\ce{\frac{n(n - 2)}{3}}$
• D. $\ce{\frac{n(n - 2)}{6}}$

Answer 1

The Correct Option is B

We have, $\frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{1+2}+\frac{1^{2}+2^{2}+3^{2}}{1+2+3}+\ldots$ $\therefore a_{n}=\frac{1^{2}+2^{2}+3^{2}+\ldots+n^{2}}{1+2+3+\ldots+n}$ $\Rightarrow a_{n}=\frac{\frac{n(n+1)(2 n+1)}{6}}{\frac{n(n+1)}{2}}$ $\Rightarrow\left[\because \Sigma n^{2}=\frac{n(n+1)(2 n+1)}{6} \text{and} \, \Sigma n=\frac{n(n+1)}{2} \therefore =\frac{2 n+1}{3}\right]$ $a_n = \frac{2n + 1}{3}$ Now, $ S_{n} =\Sigma a_{n}=\Sigma \frac{2 n+1}{3}=\frac{1}{3}\left[\sum 2 n+\Sigma 1\right] $ $=\frac{1}{3}\left[\frac{2 n(n+1)}{2}+n\right] \,\,[\because \Sigma 1=n] $ $=\frac{1}{3}\left[n^{2}+n+n\right]$ $=\frac{1}{3}\left(n^{2}+2 n\right)=\frac{n(n+2)}{3} $