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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
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Verified Answer
The correct answer is:
$\frac{n(n+2)}{3}$
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, $S_{n}=\frac{1}{3}(\Sigma n+\Sigma 1)$
$$
\begin{aligned}
&=\frac{1}{3}\left\{2 \cdot \frac{n(n+1)}{2}+n\right\}=\frac{n}{3}(n+1+1) \\
&=\frac{n(n+2)}{3}
\end{aligned}
$$
$$
\frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{1+2}+\frac{1^{2}+2^{2}+3^{2}}{1+2+3}+\ldots
$$
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, $S_{n}=\frac{1}{3}(\Sigma n+\Sigma 1)$
$$
\begin{aligned}
&=\frac{1}{3}\left\{2 \cdot \frac{n(n+1)}{2}+n\right\}=\frac{n}{3}(n+1+1) \\
&=\frac{n(n+2)}{3}
\end{aligned}
$$
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