Question

The value of $\sum _{r=1}^{\infty }\left[3\cdot 2^{-r}-2\cdot 3^{1-r}\right]$ is

• A. 2
• B. $\frac{1}{2}$
• C. 1
• D. 0

Answer 1

Answer: (D)

$\sum _{r=1}^{\infty }\left[3\cdot 2^{-r}-2\cdot 3^{1-r}\right]$
$=3\sum _{r=1}^{\infty }\frac{1}{2^r}-2\sum _{r=1}^n{3}^{1-r}$
$=3\left[\frac{1}{2}+\frac{1}{2^2}+\dots \infty \right]-2\left[1+\frac{1}{3}+\frac{1}{3^2}+\dots \infty \right]$
$=\frac{3}{2}\left[1+\frac{1}{2}+\frac{1}{2^2}+\dots \right]-2\left[1+\frac{1}{3}+\frac{1}{3^2}+\dots \infty \right]$
$=\frac{3}{2}\left(\frac{1}{1-\frac{1}{2}}\right)-2\left(\frac{1}{1-\frac{1}{3}}\right)$
$=\frac{3}{2}\times 2-\frac{2\times 3}{2}=0$