Q69 / 180 · COMEDK 2020 · Mathematics
$\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\ldots \frac{1}{(3 n-1)(3 n+2)}=$
- A$\frac{n}{6 n+3}$
- B$\frac{n}{6 n-4}$
- C$\frac{n+1}{6 n+4}$
- Correct answerD$\frac{n}{6 n+4}$
Solution
Let $S_{n}=\frac{1}{2 \times 5}+\frac{1}{5 \times 8}+\frac{1}{8 \times 11}+\ldots$
$$
\begin{aligned}
&=\frac{1}{3}\left[\frac{3}{2 \times 5}+\frac{3}{5 \times 8}+\frac{3}{8 \times 11}+\ldots+\frac{1}{(3 n-1)(3 n+2)}\right] \\
&=\frac{1}{3}\left[\frac{5-2}{2 \times 5}+\frac{8-5}{5 \times 8}+\frac{11-8}{8 \times 11}+\ldots+\frac{(3 n+2)-(3 n-1)}{(3 n-1)(3 n+2)}\right] \\
&=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\ldots+\frac{1}{3 n-1}-\frac{1}{3 n+2}\right] \\
&=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3 n+2}\right]=\frac{1}{3}\left[\frac{3 n+2-2}{2(3 n+2)}\right] \\
&=\frac{n}{2(3 n+2)}=\frac{n}{6 n+4}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.