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