Question

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

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

Answer 1

Answer: (A)

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