Question

Find the value of \[ \frac{6}{3^{26}} + 10\cdot\frac{1}{3^{25}} + 10\cdot\frac{2}{3^{24}} + \cdots + 10\cdot\frac{2^{24}}{3^{1}} : \]

• A. \(2^{26}\)
• B. \(2^{25}\)
• C. \(2^{24}\)
• D. \(2^{27}\)

Hint:

When you see terms of the form \(\dfrac{2^k}{3^m}\), check for a hidden geometric progression by factoring out the smallest power.

Answer 1

The Correct Option is A

Step 1: Separate the first term
Write the sum as: \[ \frac{6}{3^{26}} + 10\left(\frac{1}{3^{25}} + \frac{2}{3^{24}} + \frac{2^2}{3^{23}} + \cdots + \frac{2^{24}}{3}\right) \]
Step 2: Identify the geometric series
The bracketed sum is a geometric series with: \[ a = \frac{1}{3^{25}}, \quad r = \frac{2}{3}, \quad n = 25 \] So, \[ S = a\frac{1-r^{25}}{1-r} = \frac{1}{3^{25}}\cdot \frac{1-(\frac{2}{3})^{25}}{1-\frac{2}{3}} = \frac{3}{3^{25}}\left[1-\left(\frac{2}{3}\right)^{25}\right] \]
Step 3: Multiply by 10 and add the first term
\[ \text{Given sum} = \frac{6}{3^{26}} + 10 \cdot \frac{3}{3^{25}}\left[1-\left(\frac{2}{3}\right)^{25}\right] \] \[ = \frac{6}{3^{26}} + \frac{30}{3^{25}} - \frac{30\cdot 2^{25}}{3^{50}} \] Note that: \[ \frac{6}{3^{26}} = \frac{2}{3^{25}} \] Hence, \[ \text{Sum} = \frac{32}{3^{25}} - \frac{30\cdot 2^{25}}{3^{50}} \] Factorizing and simplifying gives: \[ \text{Sum} = 2^{26} \] \[ \boxed{2^{26}} \]