Question

If \( a_n = \sum_{r=0}^{n} \frac{1}{\binom{n}{r}} \), then \( \sum_{r=0}^{n} r \binom{n}{r} = \):

• A. \( (n-1) a_n \)
• B. \( n a_n \)
• C. \( \frac{n}{2} a_n \)
• D. \( a_{n+1} \)

Hint:

When dealing with sums involving binomial coefficients, use combinatorial identities to simplify the expressions. In this case, the sum \( \sum_{r=0}^{n} r \binom{n}{r} \) simplifies to \( n 2^{n-1} \), which is helpful for further manipulations.

Answer 1

The Correct Option is C

We are given: \[ a_n = \sum_{r=0}^{n} \frac{1}{\binom{n}{r}}, \] and we are asked to find the value of \( \sum_{r=0}^{n} r \binom{n}{r} \). 
Step 1: 
First, let's recall an identity from combinatorics: \[ \sum_{r=0}^{n} r \binom{n}{r} = n 2^{n-1}. \] This identity can be derived from the fact that the expression represents the sum of products of \( r \) and the binomial coefficient, which is related to the expected value of a binomial distribution.
Step 2: 
Now, observe that the given series \( a_n = \sum_{r=0}^{n} \frac{1}{\binom{n}{r}} \) can be rewritten as: \[ a_n = \sum_{r=0}^{n} \frac{1}{\binom{n}{r}} = \frac{n}{2} \cdot \sum_{r=0}^{n} r \binom{n}{r}. \] Thus, we can conclude that: \[ \sum_{r=0}^{n} r \binom{n}{r} = \frac{n}{2} a_n. \] Therefore, the correct answer is: \[ \boxed{\frac{n}{2} a_n}. \]