Q62 / 180 · KCET 2011 · Mathematics

The sum of the first $n$ terms of $\frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{1+2}+\frac{1^{2}+2^{2}+3^{2}}{1+2+3}+\ldots$ is

  • A
    $\frac{n^{2}-2 n}{3}$
  • B
    $\frac{2 n^{2}+n}{3}$
  • Correct answer
    C
    $\frac{n(n+2)}{3}$
  • D
    $\frac{2 n^{2}-n}{3}$

Solution

Given series, $$ \frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{1+2}+\frac{1^{2}+2^{2}+3^{2}}{1+2+3}+\ldots $$ The $n$th terms of the series is $$ T_{n}=\frac{\Sigma n^{2}}{\Sigma n}=\frac{n(n+1)(2 n+1)}{\frac{6 n(n+1)}{2}}=\frac{(2 n+1)}{3} $$ Now, $S_{n}=\frac{1}{3}(\Sigma n+\Sigma 1)$ $$ \begin{aligned} &=\frac{1}{3}\left\{2 \cdot \frac{n(n+1)}{2}+n\right\}=\frac{n}{3}(n+1+1) \\ &=\frac{n(n+2)}{3} \end{aligned} $$

Hence, the correct answer is option: 'C'.

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