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: , 3
Solution:
$s=1+\frac{2}{3}+\frac{7}{3^{2}}+\frac{12}{3^{3}}+\frac{17}{3^{4}}+\frac{22}{3^{5}}+\ldots$
$\frac{s}{3}=\frac{1}{3}+\frac{2}{3^{2}}+\frac{7}{3^{3}}+\ldots \ldots \infty$
$\frac{2 s}{3}=1+\frac{1}{3}+\frac{5}{3^{2}}+\frac{5}{3^{3}}+\ldots \ldots \infty$
$\frac{2 s}{3}=\frac{4}{3}+\frac{5}{3}\left\{\frac{1 / 3}{1-\frac{1}{3}}\right\}=\frac{5}{6}+\frac{4}{3}=\frac{13}{6}$
$s=\frac{13}{4}$
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