Question

Let $P=1+\frac{1}{2\times 2}+\frac{1}{2\times 2^2}+.....$ 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. $2P=3Q$
• D. $P=4Q$

Answer 1

The correct option is C $2P=3Q$

Given $P=1+\frac{1}{2\times 2}+\frac{1}{2\times 2^2}+\frac{1}{2\times 2^3}+....\infty P=\frac{1}{2}+\frac{1}{2\times 2^0}+\frac{1}{2\times 2^1}+\frac{1}{2\times 2^2}+\frac{1}{2\times 2^3}+....\infty P=\frac{1}{2}+{\sum }_{i=1}^{\infty }\frac{1}{2^n}$

Sum of infinite GP $S_{\infty }=\frac{a}{1-r}=\frac{1/2}{(1-\frac{1}{2})}=1$

$P=\frac{1}{2}+1=\frac{3}{2}Q=\frac{1}{1\times 2}+\frac{1}{3\times 4}+\frac{1}{5\times 6}+.....\infty Q={\sum }_{i=1}^{\infty }\frac{1}{n(n+1)}{T}_n=\frac{1}{n(n+1)}=[\frac{1}{n}-\frac{1}{n+1}]T_1=1-\frac{1}{2}{T}_2=\frac{1}{2}-\frac{1}{3}\underset{–}{T_n=\frac{1}{n}-\frac{1}{n+1}}{\sum }_{i=1}^n{T}_n=1-\frac{1}{n+1}=\frac{n}{n+1}n\to \infty \to \frac{1}{1+\frac{1}{n}}=1Q=1P=3/22P=3Q$