Question

The sum of the infinite series $1+\frac{2}{3}+\frac{7}{3^2}+\frac{12}{3^3}+\frac{17}{3^4}+\frac{22}{3^5}+\dots$ is equal to :

• A. $\frac{9}{4}$
• B. $\frac{15}{4}$
• C. $\frac{13}{4}$
• D. $\frac{11}{4}$

Answer 1

The correct option is C $\frac{13}{4}$

$S=1+\frac{2}{3}+\frac{7}{3^2}+\frac{12}{3^3}+\frac{17}{3^4}+\frac{22}{3^5}+\cdots \infty \dots \left(i\right)$
$\frac{S}{3}=\frac{1}{3}+\frac{2}{3^2}+\frac{7}{3^3}+\cdots \infty \dots \left(ii\right)$

From $\left(i\right)-\left(ii\right),$ we get
$\frac{2S}{3}=1+\frac{1}{3}+\frac{5}{3^2}+\frac{5}{3^3}+\cdots \infty$
$\Rightarrow \frac{2S}{3}=\frac{4}{3}+\frac{5}{3}\left(\frac{1/3}{1-\frac{1}{3}}\right)=\frac{4}{3}+\frac{5}{6}=\frac{13}{6}$
$\Rightarrow S=\frac{13}{4}$