Question

If \( C_0, C_1, C_2, \ldots, C_{15} \) are binomial coefficients in \( (1 + x)^{15} \), then \[ \frac{C_1}{C_0} + 2\frac{C_2}{C_1} + 3\frac{C_3}{C_2} + \cdots + 15\frac{C_{15}}{C_{14}} \] is equal to 

• A. \(60\)
• B. \(120\)
• C. \(64\)
• D. \(124\)
• E. \(144\)

Hint:

Use ratio identity \( \frac{C_r}{C_{r-1}} \) to simplify telescoping-like binomial sums.

Answer 1

The Correct Option is B

Concept: For binomial coefficients: \[ \frac{C_r}{C_{r-1}} = \frac{n-r+1}{r} \]

Step 1: Apply formula
\[ \frac{C_r}{C_{r-1}} = \frac{15-r+1}{r} = \frac{16-r}{r} \]

Step 2: Multiply by coefficient
\[ r \cdot \frac{C_r}{C_{r-1}} = 16-r \]

Step 3: Sum expression
\[ \sum_{r=1}^{15} (16-r) \]

Step 4: Expand
\[ = 15\cdot16 - (1+2+\cdots+15) \] \[ = 240 - \frac{15\cdot16}{2} \] \[ = 240 - 120 = 120 \]

Step 5: Final answer
\[ \boxed{120} \]