Question:

The value of $\displaystyle\sum_{r=1}^{\infty}\left[3\cdot 2^{-r} - 2\cdot 3^{1-r}\right]$ is

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Infinite GP sum: $S = \dfrac{a}{1-r}$ for $|r|<1$. Split mixed series term-by-term before summing.
Updated On: Apr 8, 2026
  • 2
  • $1/2$
  • 1
  • 0
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
Split the series and use the formula for sum of an infinite geometric progression.
Step 2: Detailed Explanation:
$\displaystyle\sum_{r=1}^\infty 3 \cdot 2^{-r} = 3 \cdot \frac{1/2}{1-1/2} = 3$.
$\displaystyle\sum_{r=1}^\infty 2 \cdot 3^{1-r} = 2 \cdot 3^0 + 2\cdot3^{-1}+\cdots = 2 \cdot \frac{1}{1-1/3} = 3$.
Difference $= 3 - 3 = 0$.
Step 3: Final Answer:
The sum is $0$.
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