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

Answer 1

Answer: (A)

$S=1+\frac{2}{3}+\frac{7}{3^2}+\frac{12}{3^3}+\frac{17}{3^4}+\dots$
$\frac{S}{3}=\frac{1}{3}+\frac{2}{3^2}+\frac{7}{3^3}+\frac{12}{3^4}+\dots$
$\frac{2S}{3}=1+\frac{1}{3}+\frac{5}{3^2}+\frac{5}{3^3}+\frac{5}{3^4}+\dots +$ up to infinite terms
$\Rightarrow S=\frac{13}{4}$