Q50 / 160 · TS EAMCET 2013 · 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
    $\log \left(\frac{2}{e}\right)$
  • Correct answer
    B
    $\log \left(\frac{e}{2}\right)$
  • C
    $\log (2 e)$
  • D
    $e-1$

Solution

$\begin{aligned} & \frac{1}{2 \cdot 3}+\frac{1}{4 \cdot 5}+\frac{1}{6 \cdot 7}+\frac{1}{8 \cdot 9}+\ldots \infty \\ & =\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{8}-\frac{1}{9}\right)+\ldots \infty \\ & =1-\log _e 2 \\ & =\log _e e-\log _e 2 \\ & =\log \left(\frac{e}{2}\right)\end{aligned}$

Hence, the correct answer is option: 'B'.

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