\; {6}{3²6}} + \; {10 · 1}{3²5}} + \; {10 · 2}{3²4}} + \; {10 · 2²}{3²3}} + + \; {10 · 2²4}}{3} is equal to : [28 Jan 2026 Shift 2]

Correct Answer is : 2262²6}226

Correct Answer is : 2262²6}226

Test Index

Sequences and Series Part 3

Section: Mathematics

Question : 17 of 109

Marks: +1, -0

\; \frac{6}{3^{26}} + \; \frac{10 \cdot 1}{3^{25}} + \; \frac{10 \cdot 2}{3^{24}} + \; \frac{10 \cdot 2^{2}}{3^{23}} + \dots + \; \frac{10 \cdot 2^{24}}{3}

Solution:

Concept:

The series is a geometric progression after the first term.

We separate the first term and sum the remaining GP using the formula

S_n = a \frac{r^n - 1}{r - 1}

Explanation:

Let

S = \frac{6}{3^{26}} + \frac{10 \cdot 1}{3^{25}} + \frac{10 \cdot 2}{3^{24}} + \frac{10 \cdot 2^2}{3^{23}} + \dots + \frac{10 \cdot 2^{24}}{3}

From the second term onward, factor

\frac{10}{3^{25}}

S' = \frac{10}{3^{25}} \left( 2^0 \cdot 3^0 + 2^1 \cdot 3^1 + 2^2 \cdot 3^2 + \dots + 2^{24} \cdot 3^{24} \right)

Each term is

(2 \cdot 3)^k = 6^k

for

k=0

24

. So

S' = \frac{10}{3^{25}} \sum_{k=0}^{24} 6^k

This is a GP with

a=1

r=6

n=25

. Its sum is

\frac{6^{25} - 1}{6 - 1} = \frac{6^{25} - 1}{5}

Thus

S' = \frac{10}{3^{25}} \cdot \frac{6^{25} - 1}{5} = \frac{2}{3^{25}} (6^{25} - 1)

Since

6^{25} = (2 \cdot 3)^{25} = 2^{25} \cdot 3^{25}

, we get

S' = \frac{2 \cdot 2^{25} \cdot 3^{25}}{3^{25}} - \frac{2}{3^{25}} = 2^{26} - \frac{2}{3^{25}}

Add the first term:

\frac{6}{3^{26}} = \frac{2 \cdot 3}{3^{26}} = \frac{2}{3^{25}}

Hence

S = \frac{2}{3^{25}} + 2^{26} - \frac{2}{3^{25}} = 2^{26}

Answer:

2^{26}

, which corresponds to option A.

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