KCET · Maths · Sequences and Series
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 \(\frac{2 n^{2}+n}{3}\)
- C \(\frac{n(n+2)}{3}\)
- D \(\frac{2 n^{2}-n}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{n(n+2)}{3}\)
Step-by-step Solution
Detailed explanation
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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