Q107 / 180 · COMEDK 2024 (Memory Based) · Mathematics
The value of $\frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\frac{1}{7 \cdot 10}+\ldots+\frac{1}{(3 n-2)(3 n+1)}$ is
- Correct answerA$\frac{n}{3 n+1}$
- B$\frac{n}{3 n+2}$
- C$\frac{-n}{3 n+1}$
- D$\frac{-n}{3 n+2}$
Solution
$\begin{aligned} & \text { } \frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\frac{1}{7 \cdot 10}+\ldots+\frac{1}{(3 n-2)(3 n+1)} \\ & \begin{aligned} & \frac{1}{3}\left[\frac{4-1}{1 \cdot 4}+\frac{7-4}{4 \cdot 7}+\frac{10-7}{7 \cdot 10}+\ldots+\frac{(3 n+1)-(3 n-2)}{(3 n-2)(3 n+1)}\right] \\ = & \frac{1}{3}\left[\frac{4}{1 \cdot 4}-\frac{1}{1 \cdot 4}+\frac{7}{4 \cdot 7}-\frac{4}{4 \cdot 7}+\frac{10}{7 \cdot 10}-\frac{7}{7 \cdot 10}\right. \\ & \left.+\ldots+\frac{1}{3 n-2}-\frac{1}{3 n+1}\right] \\ = & \frac{1}{3}\left[\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\ldots+\frac{1}{3 n-2}-\frac{1}{3 n+1}\right] \\ = & \frac{1}{3}\left[1-\frac{1}{3 n+1}\right]=\frac{3 n}{3(3 n+1)}=\frac{n}{3 n+1}\end{aligned}\end{aligned}$
Hence, the correct answer is option: 'A'.