AP EAMCET · Maths · Sequences and Series
\(\frac{1}{1 \cdot 3}+\frac{1}{2 \cdot 5}+\frac{1}{3 \cdot 7}+\frac{1}{4 \cdot 9}+\ldots\) is equal to
- A \(2 \log _e 2-2\)
- B \(2-\log _e 2\)
- C \(2 \log _e 4\)
- D \(\log _e 4\)
Answer & Solution
Correct Answer
(B) \(2-\log _e 2\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Let } \quad S=\frac{1}{1 \cdot 3}+\frac{1}{2 \cdot 5}+\frac{1}{3 \cdot 7}+\frac{1}{4 \cdot 9}+\ldots \\ & \therefore \quad T_n=\frac{1}{n(2 n+1)} \\ & =\frac{1}{n}-\frac{2}{(2 n+1)} \\ & \Rightarrow \quad S=\sum_{n=1}^{\infty}…
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