Question

Evaluate: $\sum _{i=1}^{\infty }\frac{2i-1}{2^{i+1}}$

• A. $1.5$
• B. $2.5$
• C. $2.0$
• D. None of these

Answer 1

The correct option is A $1.5$

Let the given summation be $L$

$L=\sum _{i=1}^{\infty }\frac{2i-1}{2^{i+1}}=\sum _{i=1}^{\infty }\frac{i}{2^i}-\sum _{i=1}^{\infty }\frac{1}{2^{i+1}}$

Consider $\sum _{i=1}^{\infty }\frac{i}{2^i}=1.\frac{1}{2}+2.\frac{1}{2^2}+3.\frac{1}{2^3}+.......\infty =\frac{1}{2}(1+2.\frac{1}{2}+3.\frac{1}{2^2}+.....+\infty )$

Observe that, it is a arithmetic geometric progression with $a=1,d=1,r=\frac{1}{2}$

$\Rightarrow \sum _{i=1}^{\infty }\frac{i}{2^i}=\frac{1}{2}(\frac{a}{1-r}+\frac{rd}{(1-r{)}^2})=\frac{1}{2}(2+2)=2$

Now consider $\sum _{i=1}^{\infty }\frac{1}{2^{i+1}}=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+.....+\infty =\frac{1}{2^2}(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\infty )$

Observe that, it is a geometric progression with $a=1,r=\frac{1}{2}$

$\Rightarrow \sum _{i=1}^{\infty }\frac{1}{2^{i+1}}=\frac{1}{4}\left(\frac{a}{1-r}\right)=\frac{1}{4}\left(2\right)=\frac{1}{2}$

Therefore the value of $L=\sum _{i=1}^{\infty }\frac{i}{2^i}-\sum _{i=1}^{\infty }\frac{1}{2^{i+1}}=2-0.5=1.5$

Hence, option $A$ is correct.