KCET · Maths · Sequences and Series
The sum of \( 1^{s t} \mathrm{n} \) terms of the series
\[
\frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{1+2}+\frac{1^{2}+2^{2}+3^{2}}{1+2+3}+\ldots
\]
- A \( \frac{n+2}{3} \)
- B \( \frac{n(n+2)}{3} \)
- C \( \frac{n(n-2)}{3} \)
- D \( \frac{n(n-2)}{6} \)
Answer & Solution
Correct Answer
(B) \( \frac{n(n+2)}{3} \)
Step-by-step Solution
Detailed explanation
Given that,
\[
\frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{1+2}+\frac{1^{2}+2^{2}+3^{2}}{1+2+3}+\cdot s
\]
So, nth term is given by
\[
\begin{array}{l}
t_{n}=\frac{1^{2}+2^{2}+\ldots+n^{2}}{1+2+\ldots+n} \\
=\frac{n(n+1)(2 n+1)}{6} \times \frac{2}{n(n+1)} \\
=\frac{(2 n+1)}{3}
\end{array}
\]
\[
\begin{array}{l}
\text { Now, } \Sigma t_{n}=\frac{1}{3}(2 \Sigma n+\Sigma 1) \\
=\frac{1}{3}\left[\frac{2 \times n(n+1)}{2}+n\right]=\frac{1}{3} n(n+2)
\end{array}
\]
\[
\frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{1+2}+\frac{1^{2}+2^{2}+3^{2}}{1+2+3}+\cdot s
\]
So, nth term is given by
\[
\begin{array}{l}
t_{n}=\frac{1^{2}+2^{2}+\ldots+n^{2}}{1+2+\ldots+n} \\
=\frac{n(n+1)(2 n+1)}{6} \times \frac{2}{n(n+1)} \\
=\frac{(2 n+1)}{3}
\end{array}
\]
\[
\begin{array}{l}
\text { Now, } \Sigma t_{n}=\frac{1}{3}(2 \Sigma n+\Sigma 1) \\
=\frac{1}{3}\left[\frac{2 \times n(n+1)}{2}+n\right]=\frac{1}{3} n(n+2)
\end{array}
\]
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