Question

If $P=1+\frac{1}{2\times 2}+\frac{1}{3\times 2^2}+...$ and $Q=\frac{1}{1\times 2}+\frac{1}{3\times 4}+\frac{1}{5\times 6}+.....$, then

• A. $P=Q$
• B. $2P=Q$
• C. $P=2Q$
• D. $P=4Q$

Answer 1

The correct option is C $P=2Q$

Given $P=1+\frac{1}{2\times 2}+\frac{1}{3\times 2^2}+....$
and

$Q=\frac{1}{1\times 2}+\frac{1}{3\times 4}+\frac{1}{5\times 6}+.....$

Now,

$\frac{P}{2}=\frac{{\left(1/2\right)}^1}{1}+\frac{{\left(1/2\right)}^2}{2}+\frac{{\left(1/2\right)}^3}{3}+...$

$\Rightarrow -\frac{P}{2}=-(\frac{{\left(1/2\right)}^1}{1}-\frac{{\left(1/2\right)}^2}{2}-\frac{{\left(1/2\right)}^3}{3}-...)$

$\Rightarrow -\frac{P}{2}={\log }_e(1-\frac{1}{2})$

$\Rightarrow -\frac{P}{2}={\log }_e\frac{1}{2}$

$\Rightarrow P=2{\log }_{e}2....\left(i\right)$

Now,

$Q=(1-\frac{1}{2})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{5}-\frac{1}{6})+....$

$\Rightarrow Q={\log }_{e}2....\left(ii\right)$

Now, from eqs. $\left(i\right)$ and $\left(ii\right)$, we get

$P=2Q$