Q118 / 200 · WBJEE 2013 · Mathematics
If $P=1+\frac{1}{2 \times 2}+\frac{1}{3 \times 2^{2}}+\ldots$
and $Q=\frac{1}{1 \times 2}+\frac{1}{3 \times 4}+\frac{1}{5 \times 6}+\ldots,$ then
- A$P=Q$
- B$2 P=Q$
- Correct answerC$P=2 Q$
- D$P=4 Q$
Solution
Given,
$$
P=1+\frac{1}{2 \times 2}+\frac{1}{3 \times 2^{2}}+\ldots
$$
and $\quad Q=\frac{1}{1 \times 2}+\frac{1}{3 \times 4}+\frac{1}{5 \times 6}+\ldots$
Now, $\quad \frac{P}{2}=\frac{(1 / 2)^{1}}{1}+\frac{(1 / 2)^{2}}{2}+\frac{(1 / 2)^{3}}{3}+\ldots$
$\Rightarrow \quad-p / 2=-\left(\frac{1 / 2}{1}\right)^{1}-\frac{(1 / 2)^{2}}{3}-\frac{(1 / 2)^{3}}{3} \ldots$
$\Rightarrow \quad-P / 2=\log _{e}\left(1-\frac{1}{2}\right)$
$\Rightarrow \quad-\frac{P}{2}=\log _{\theta} \frac{1}{2}$
$\Rightarrow \quad P=2 \log _{e} 2$
Now, $Q=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\ldots$
$\Rightarrow \quad Q=\log _{e} 2$
Now, from Eqs. (i) and (ii), we get
$$
P=2 Q
$$
Hence, the correct answer is option: 'C'.