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

Concept: This problem utilizes the symmetry of binomial coefficients, \( ^nC_k = ^nC_{n-k} \). By substituting \( k \) with \( (n-k) \) in a summation, we can often simplify complex expressions.

Step 1: Let the given sum be \( S \).
\[ S = \sum_{k=0}^{18} \frac{k}{^{18}C_k} \quad \cdots (1) \] Using the property \( ^{18}C_k = ^{18}C_{18-k} \), we can rewrite the sum by replacing \( k \) with \( 18-k \): \[ S = \sum_{k=0}^{18} \frac{18-k}{^{18}C_{18-k}} = \sum_{k=0}^{18} \frac{18-k}{^{18}C_k} \quad \cdots (2) \]

Step 2: Add equations (1) and (2).
\[ 2S = \sum_{k=0}^{18} \frac{k}{^{18}C_k} + \sum_{k=0}^{18} \frac{18-k}{^{18}C_k} \] \[ 2S = \sum_{k=0}^{18} \frac{k + 18 - k}{^{18}C_k} = \sum_{k=0}^{18} \frac{18}{^{18}C_k} \] \[ 2S = 18 \sum_{k=0}^{18} \frac{1}{^{18}C_k} \]

Step 3: Find the value of \( a \).
Dividing both sides by 2: \[ S = 9 \sum_{k=0}^{18} \frac{1}{^{18}C_k} \] Comparing this with the original equation \( S = a \sum_{k=0}^{18} \frac{1}{^{18}C_k} \), we see that: \[ a = 9 \]