Binomial TheoremHard

Question

$\frac{6}{3^{26}} + \frac{10.1}{3^{25}} + \frac{10.2}{3^{24}} + \frac{{10.2}^{2}}{3^{23}} + \ldots + \frac{{10.2}^{24}}{3}$ is equal to

Options

A.$2^{25}$
B.$2^{26}$
C.$3^{25}$
D.$3^{26}$

Solution

$\ S = \frac{6}{3^{26}} + \frac{10}{3^{25}}\left\lbrack \frac{(6)^{25} - 1}{6 - 1} \right\rbrack$

$$\begin{matrix} & S = \frac{6}{3^{26}} + \frac{10}{3^{25}}\left\lbrack \frac{6^{25} - 1}{5} \right\rbrack \\ & S = \frac{2}{3^{25}} + 2\left\lbrack 2^{25} - \frac{1}{3^{25}} \right\rbrack \\ & S = 2^{26} \end{matrix}$$

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