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Manipal Entrance Test 2015 Math Paper

Question : 67 of 70

Marks: +1, -0

1+\frac{2}{3}+\frac{6}{3^{2}}

+\frac{10}{3^{3}}+\frac{14}{3^{4}}+\dots

Solution:

Let S be the sum of the given series.

i.e.

S=1+\frac{2}{3}+\frac{6}{3^{2}}

+\frac{10}{3^{3}}+\frac{14}{3^{4}}+\dots

\Rightarrow (S-1)=\frac{2}{3}+\frac{6}{3^{2}}+\frac{10}{3^{3}}+\dots\;\;\;

...(i)

\Rightarrow (S-1) \times \frac{1}{3}= \frac{2}{3^{2}}+\frac{6}{3^{3}}+\frac{10}{3^{4}}+\dots\;\;\;

...(ii)

On subtracting Eq. (ii) from Eq. (i), we get

\frac{2}{3}(S-1)

= \frac{2}{3}+\frac{4}{3^{2}}+\frac{4}{3^{3}}+\frac{4}{3^{4}}+\dots

\Rightarrow \frac{2}{3}(S-1)= \frac{2}{3}+ \frac{\frac{4}{3^{2}}}{1-\frac{1}{3}}

\Rightarrow \frac{2}{3}(S-1)= \frac{2}{3}+ \frac{2}{3}

\Rightarrow S=3

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