Question ID: #971
$\frac{6}{3^{26}}+\frac{10 \cdot 1}{3^{25}}+\frac{10 \cdot 2}{3^{24}}+\frac{10 \cdot 2^{2}}{3^{23}}+…+\frac{10 \cdot 2^{24}}{3}$ is equal to
- (1) $2^{25}$
- (2) $2^{26}$
- (3) $3^{25}$
- (4) $3^{26}$
Solution:
Let the given series be $S$.
$$S = \frac{6}{3^{26}} + \left[ \frac{10}{3^{25}} + \frac{10 \cdot 2}{3^{24}} + \frac{10 \cdot 2^2}{3^{23}} + \dots + \frac{10 \cdot 2^{24}}{3} \right]$$
Observe the terms inside the bracket. They form a Geometric Progression (G.P.).
First term $A = \frac{10}{3^{25}}$.
Second term $A_2 = \frac{10 \cdot 2}{3^{24}}$.
Common ratio $R = \frac{A_2}{A} = \frac{10 \cdot 2 / 3^{24}}{10 / 3^{25}} = 2 \cdot 3 = 6$.
The terms range from $2^0$ to $2^{24}$, so there are $n = 25$ terms in this G.P.
Formula for the sum of a G.P. is $S_n = A \frac{R^n – 1}{R – 1}$.
$$S_{25} = \frac{10}{3^{25}} \left[ \frac{6^{25} – 1}{6 – 1} \right]$$
$$S_{25} = \frac{10}{3^{25}} \left[ \frac{6^{25} – 1}{5} \right]$$
$$S_{25} = \frac{2}{3^{25}} (6^{25} – 1)$$
Now substitute this back into the original sum $S$:
$$S = \frac{6}{3^{26}} + \frac{2}{3^{25}} (6^{25} – 1)$$
Simplify the first term:
$$\frac{6}{3^{26}} = \frac{2 \cdot 3}{3^{26}} = \frac{2}{3^{25}}$$
Substitute and expand:
$$S = \frac{2}{3^{25}} + \frac{2 \cdot 6^{25}}{3^{25}} – \frac{2}{3^{25}}$$
$$S = \frac{2 \cdot 6^{25}}{3^{25}}$$
$$S = 2 \left( \frac{6}{3} \right)^{25}$$
$$S = 2 (2^{25})$$
$$S = 2^{26}$$
Ans. (2)
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