Question

If $C_0, C_1, C_2, \dots, C_n$ are the binomial coefficients in the expansion of $(1+x)^n$ then $\sum_{r=1}^{n} \frac{r C_r}{C_{r-1}} =$

• A. 540
• B. 336
• C. 105
• D. 270

Hint:

When working with sums of binomial coefficients, the ratio $\frac{C_r}{C_{r-1}} = \frac{n-r+1}{r}$ is the most fundamental tool for simplification. If you arrive at a formula that depends on $n$ but the options are constants, check if the context implies a specific value for $n$. If not, and no integer $n$ satisfies the condition, the question is likely erroneous.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept: The ratio of consecutive binomial coefficients is given by the formula: \[ \frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r} \] In this problem, \( n = 8 \).
Step 2: Simplify the General Term: Let the general term be \( T_r \). \[ T_r = r^3 \cdot \frac{C_r}{C_{r-1}} = r^3 \cdot \frac{8-r+1}{r} \] \[ T_r = r^3 \cdot \frac{9-r}{r} = r^2(9-r) = 9r^2 - r^3 \]
Step 3: Calculating the Sum: We need to find \( S = \sum_{r=1}^{8} (9r^2 - r^3) \). Using summation formulas for the first \( n \) natural numbers: 1. \( \sum r^2 = \frac{n(n+1)(2n+1)}{6} \) 2. \( \sum r^3 = \left[\frac{n(n+1)}{2}\right]^2 \) For \( n=8 \): \[ \sum_{r=1}^{8} r^2 = \frac{8(8+1)(16+1)}{6} = \frac{8 \cdot 9 \cdot 17}{6} = 12 \cdot 17 = 204 \] \[ \sum_{r=1}^{8} r^3 = \left[\frac{8(8+1)}{2}\right]^2 = (36)^2 = 1296 \] Substituting these values into the sum expression: \[ S = 9(204) - 1296 \] \[ S = 1836 - 1296 = 540 \]
Step 4: Final Answer: The sum is 540.