Question

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

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

Answer 1

The correct option is C $P=2Q$

We know $\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-...-\log (1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}-...$

$Q=\frac{1}{1\times 2}+\frac{1}{3\times 4}+\frac{1}{5\times 6}+....\infty =\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....\infty Q=\log (1+1)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...Q=\log 2-\log (1-\frac{1}{x})=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...$

Putting $x=\frac{1}{2}$

$-\log (1-\frac{1}{2})=\frac{1}{2}+\frac{1}{2\times 2^2}+\frac{1}{3\times 2^3}+\frac{1}{4\times 2^4}+...-2\log \left(\frac{1}{2}\right)=1+\frac{1}{2\times 2}+\frac{1}{3\times 2^2}+\frac{1}{4\times 2^3}+....P=2\log 2=1+\frac{1}{2\times 2}+\frac{1}{3\times 2^2}+\frac{1}{4\times 2^3}+....\frac{P}{Q}=2P=2Q$