The sum of the infinite series

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

  1. $\frac{13}{4}$

  2. $\frac{9}{4}$

  3. $\frac{15}{4}$

  4. $\frac{11}{4}$


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

$\Rightarrow \mathrm{S}=\frac{13}{4}$

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