Question

If $\sum_{k=0}^{18} \frac{k}{\binom{18}{k}} = a \sum_{k=0}^{18} \frac{1}{\binom{18}{k}}$, then the value of $a$ is equal to:

• A. \( 3 \)
• B. \( 9 \)
• C. \( 6 \)
• D. \( 18 \)
• E. \( 36 \)

Hint:

For any sum of the form \( \sum_{k=0}^{n} \frac{k}{^nC_k} \), the result is always \( \frac{n}{2} \sum_{k=0}^{n} \frac{1}{^nC_k} \). Here, \( n=18 \), so \( a = 18/2 = 9 \).

Answer 1

The Correct Option is B