Question

Value of \[ \sum_{k=1}^{\infty}\sum_{r=0}^{k}\frac{1}{3^{k}}\binom{k}{r} \] is:

• A. \(1\)
• B. \(0\)
• C. \(\dfrac{2}{3}\)
• D. \(2\)

Hint:

Remember the binomial identity: \[ \sum_{r=0}^{k}\binom{k}{r}=2^k \] This often converts double summations into simple geometric series.

Answer 1

The Correct Option is D

Step 1: Evaluate the inner summation. Using the identity: \[ \sum_{r=0}^{k} \binom{k}{r} = 2^k \] So the given expression becomes: \[ \sum_{k=1}^{\infty} \frac{1}{3^k}\,2^k \]
Step 2: Simplify the series. \[ \sum_{k=1}^{\infty} \left(\frac{2}{3}\right)^k \] This is a geometric series with: \[ a=\frac{2}{3}, \qquad r=\frac{2}{3} \]
Step 3: Use the sum formula of an infinite GP. \[ S=\frac{a}{1-r} =\frac{\frac{2}{3}}{1-\frac{2}{3}} =\frac{\frac{2}{3}}{\frac{1}{3}} =2 \]
Hence, the required value is \[ \boxed{2} \]