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Army Group X Practice Set 3

Question : 55 of 70

Marks: +1, -0

1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\cdots\infty

,is

Solution:

Let,

S=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\cdots\infty \ldots(i)

Multiplying by

\frac{1}{3}

\frac{1}{3}S=\frac{1}{3}+\frac{2}{3^2}+\frac{6}{3^3}+\frac{10}{3^4}+\frac{14}{3^5}+\cdots\infty \ldots(ii)

Subtracting Eq. (ii) from Eq. (i),

\frac{2}{3}S=1+\frac{1}{3}+\left(\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\cdots\infty\right)

=1+\frac{1}{3}+\frac{\frac{4}{3^2}}{1-\frac{1}{3}}

\left[\because \text{In GP, } a=\frac{4}{3^2}, r=\frac{1}{3} \text{ and } \sum \text{ of infinite terms of GP } = \frac{a}{1-r}\right]

=\frac{4}{3}+\frac{4}{9}\times\frac{3}{2}

=\frac{4}{3}+\frac{2}{3}=\frac{6}{3}=2

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