TS EAMCET · Maths · Basic of Mathematics
\(\frac{1}{2 \cdot 3}+\frac{1}{4 \cdot 5}+\frac{1}{6 \cdot 7}+\frac{1}{8 \cdot 9}+\ldots\) is equal to
- A \(\log \left(\frac{2}{e}\right)\)
- B \(\log \left(\frac{e}{2}\right)\)
- C \(\log (2 e)\)
- D \(e-1\)
Answer & Solution
Correct Answer
(B) \(\log \left(\frac{e}{2}\right)\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \frac{1}{2 \cdot 3}+\frac{1}{4 \cdot 5}+\frac{1}{6 \cdot 7}+\frac{1}{8 \cdot 9}+\ldots \infty \\ & =\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{8}-\frac{1}{9}\right)+\ldots \infty…
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