Binomial Theorem
Double summation
Grade 11
Question:
<p>Value of \(\displaystyle\sum_{k=1}^{\infty}\sum_{r=0}^{k} \dfrac{1}{3^k}\binom{k}{r}\) is</p>
<p>(1) \(\dfrac{2}{3}\)</p>
<p>(2) \(\dfrac{4}{3}\)</p>
<p>(3) 2</p>
<p>(4) 1</p>
Step-by-Step Solution
Key Concept: Recognize that the inner sum ∑(k choose r) from r=0 to k equals 2^k by the binomial theorem. This transforms the double sum into a geometric series that can be evaluated.
<p><strong>Step 1:</strong> Evaluate the inner sum using the binomial theorem.</p><p>By the binomial theorem: $\sum_{r=0}^{k}\binom{k}{r} = 2^k$</p><p><strong>Step 2:</strong> Substitute into the double sum.</p><p>$$\sum_{k=1}^{\infty}\sum_{r=0}^{k} \frac{1}{3^k}\binom{k}{r} = \sum_{k=1}^{\infty} \frac{1}{3^k} \cdot 2^k = \sum_{k=1}^{\infty}\left(\frac{2}{3}\right)^k$$</p><p><strong>Step 3:</strong> Evaluate the geometric series with first term a = 2/3 and common ratio r = 2/3.</p><p>$$\sum_{k=1}^{\infty}\left(\frac{2}{3}\right)^k = \frac{\frac{2}{3}}{1-\frac{2}{3}} = \frac{\frac{2}{3}}{\frac{1}{3}} = 2$$</p><p>∴ Answer: C (which equals 2)</p>
Correct Answer: C