Q65 / 110 · VITEEE 2024 · Mathematics
Sum of first ' $n$ ' terms of a series $a_1+a_2+\ldots+a_n$ is given by $S_n=\frac{n\left(n^2-1\right)(n+2)}{4}$, then the value of $\lim _{n \rightarrow \infty} \sum_{r=2}^n \frac{1}{a_r}$ is
- A4
- B2
- Correct answerC$\frac{1}{4}$
- D$\frac{1}{2}$
Solution
$\text { } \begin{aligned}
S_n & =\frac{n\left(n^2-1\right)(n+2)}{4} \\
& =\frac{n(n+1)(n-1)(n+2)}{4} \\
& =\frac{n(n+1)}{4}\left(n^2+n-2\right) \\
\Rightarrow \quad S_n & =\frac{n(n+1)}{4}\left(n^2+n\right)-\frac{n(n+1)}{4}(2) \\
& =\frac{n^2(n+1)^2}{4}-\frac{n(n+1)}{2} \\
\because \quad \frac{n^2(n+1)^2}{4} & =\Sigma n^3 \text { and } \frac{n(n+1)}{4}=\Sigma n \\
\Rightarrow \quad S_n & =\Sigma\left(n^3-n\right)
\end{aligned}$
Therefore, $a_n=n^3-n$
$\Rightarrow a_r=r^3-r=r\left(r^2-1\right)=r(r-1)(r+1)$
and $\sum_{r=2}^n \frac{1}{a_r}=\sum_{r=2}^n \frac{1}{r(r+1)(r-1)}=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{n(n+1)}\right)$ and $\lim _{n \rightarrow \infty} \sum_{r=2}^n \frac{1}{a}=\frac{1}{2}\left(\frac{1}{2}-0\right)=\frac{1}{4}$
Hence, the correct answer is option: 'C'.