Question

If the sum to infinity of the series $1+2r+{3r}^2+{4r}^3+\cdot \cdot \cdot$ is $\frac{9}{4}$, then the value of $r$ is

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. none of these

Answer 1

Answer: (B)

$\frac{1}{3}$

$\frac{9}{4}=1+2r+3r^2+4r^3\cdots \to \left(1\right)$
By mutiplying above equation by $r$, we get,
$\frac{9}{4}r=r+2r^2+3r^3+4r^4\cdots \to \left(2\right)$
Subtracting (2) from (1) by shifting one place, we get
$\frac{9}{4}(1-r)=1+r{+r}^2+r^3+r^4.......$
The above equation becomes ,
$\frac{9}{4}(1-r)=\frac{1}{1-r}$
${(1-r)}^2=\frac{4}{9}$
Hence, we get
$r=\frac{1}{3}$
(since $r$ is positive)