Question

If \( C_j \) stands for \( ^nC_j \), then \[ \frac{C_0}{2} + \frac{C_1}{2.2^2} + \frac{C_2}{3.2^3} + \dots + C_n = \frac{3^n}{2^{n+1} (n+1)} \]

• A. \( \frac{3^n}{2^{n+1} (n+1)} \)
• B. \( \frac{3^{n+1}}{2^{n+1} (n+1)} \)
• C. \( \frac{3^n}{2^n (n+1)} \)
• D. \( \frac{3^{n+1}}{2^n (n+1)} \)

Hint:

The sum of binomial coefficients is related to the expansion of \( (1+1)^n \), and this can be used to derive expressions for more complex series sums.

Answer 1

The Correct Option is B

We know that \( C_j \) stands for the binomial coefficients: \[ C_j = \binom{n}{j} \] The sum of all binomial coefficients for a fixed \( n \) is known to be: \[ \sum_{j=0}^{n} \binom{n}{j} = 2^n \] Thus, the sum of the given series is: \[ C_0 + C_1 + C_2 + \dots + C_n = 2^n \] After applying the given transformation, we simplify the expression, leading to the final result: \[ \frac{3^{n+1}}{2^{n+1} (n+1)} \] Thus, the correct answer is \( \frac{3^{n+1}}{2^{n+1} (n+1)} \).