Q97 / 180 · KCET 2016 · Mathematics
The sum of \( 1^{s t} \mathrm{n} \) terms of the series
\[
\frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{1+2}+\frac{1^{2}+2^{2}+3^{2}}{1+2+3}+\ldots
\]
\[ \frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{1+2}+\frac{1^{2}+2^{2}+3^{2}}{1+2+3}+\ldots \]
- A\( \frac{n+2}{3} \)
- Correct answerB\( \frac{n(n+2)}{3} \)
- C\( \frac{n(n-2)}{3} \)
- D\( \frac{n(n-2)}{6} \)
Solution
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
\[ \begin{array}{l} t_{n}=\frac{1^{2}+2^{2}+\ldots+n^{2}}{1+2+\ldots+n} \\ =\frac{n(n+1)(2 n+1)}{6} \times \frac{2}{n(n+1)} \\ =\frac{(2 n+1)}{3} \end{array} \]
\[ \begin{array}{l} \text { Now, } \Sigma t_{n}=\frac{1}{3}(2 \Sigma n+\Sigma 1) \\ =\frac{1}{3}\left[\frac{2 \times n(n+1)}{2}+n\right]=\frac{1}{3} n(n+2) \end{array} \]
Hence, the correct answer is option: 'B'.