COMEDK · Maths · 5. Sequences and Series
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
- A \(\frac{n}{3 n+1}\)
- B \(\frac{n}{3 n+2}\)
- C \(\frac{-n}{3 n+1}\)
- D \(\frac{-n}{3 n+2}\)
Answer & Solution
Correct Answer
(A) \(\frac{n}{3 n+1}\)
Step-by-step Solution
Detailed explanation
\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…
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