frac{6}{3^{26}} + frac{10 cdot 1}{3^{25}} + f | vedclass

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Question

(B) Let the given expression be $S = \frac{6}{3^{26}} + \sum_{k=0}^{24} \frac{10 \cdot 2^k}{3^{25-k}}$.This can be rewritten as $S = \frac{6}{3^{26}}

Vedclass · Accepted Answer

$2^{26}$

Vedclass · Answer

(B) Let the given expression be $S = \frac{6}{3^{26}} + \sum_{k=0}^{24} \frac{10 \cdot 2^k}{3^{25-k}}$.This can be rewritten as $S = \frac{6}{3^{26}} + \frac{10}{3^{25}} \sum_{k=0}^{24} \left(\frac{2}{3^{-1}}\right)^k = \frac{6}{3^{26}} + \frac{10}{3^{25}} \sum_{k=0}^{24} (6)^k$.Using the sum formula for a geometric progression $\sum_{k=0}^{n-1} r^k = \frac{r^n - 1}{r - 1}$,we get:$S = \frac{6}{3^{26}} + \frac{10}{3^{25}} \left[ \frac{6^{25} - 1}{6 - 1} \right] = \frac{6}{3^{26}} + \frac{10}{3^{25}} \left[ \frac{6^{25} - 1}{5} \right]$.$S = \frac{6}{3^{26}} + \frac{2}{3^{25}} (6^{25} - 1) = \frac{6}{3^{26}} + \frac{2 \cdot 6^{25}}{3^{25}} - \frac{2}{3^{25}}$.Since $6^{25} = 2^{25} \cdot 3^{25}$,we have $S = \frac{6}{3^{26}} + 2 \cdot 2^{25} - \frac{2}{3^{25}} = \frac{2 \cdot 3}{3^{26}} + 2^{26} - \frac{2}{3^{25}} = \frac{2}{3^{25}} + 2^{26} - \frac{2}{3^{25}} = 2^{26}$.

$\frac{6}{3^{26}} + \frac{10 \cdot 1}{3^{25}} + \frac{10 \cdot 2}{3^{24}} + \frac{10 \cdot 2^2}{3^{23}} + \ldots + \frac{10 \cdot 2^{24}}{3}$ is equal to

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