Question

If \( C_j \) stands for \( \binom{n}{j} \), then: \[ \frac{C_1}{C_0} + 2 \times \frac{C_2}{C_1} + 3 \times \frac{C_3}{C_2} + ... + n \times \frac{C_n}{C_{n-1}} \]

• A. $\sum_{k=1}^{n} k^2 $
• B. $\sum_{k=1}^{n} k^3 $
• C. $\sum_{k=1}^{n} \frac{k}{2} $
• D. $\sum_{k=1}^{n} k $

Hint:

Use properties of binomial coefficients and their sums for simplification. The weighted sum can often be simplified using binomial expansion identities.

Answer 1

The Correct Option is D

The given sum can be interpreted as the weighted sum of binomial coefficients. Using combinatorial identities, we find that the sum simplifies to: \[ \sum_{k=1}^{n} k = n \times 2^{n-1} \] Thus, the correct answer is option (4).