Question

If \( \sum_{r=0}^5 \frac{{}^{11}C_{2r+1}}{2r+2} = \frac{m}{n} \), gcd(m, n) = 1, then \( m - n \) is equal to:

Hint:

For summations involving binomial coefficients, integrating the binomial expansion and using symmetry can simplify the problem significantly.

Answer 1

Correct Answer: 2035

Solution for Binomial Sum Problem

The problem requires calculating the sum:

\[ \sum_{r=0}^5 \frac{{}^{11}C_{2r+1}}{2r+2} \] The general term is \( \frac{{}^{11}C_{2r+1}}{2r+2} \), where \( {}^{11}C_k \) denotes the binomial coefficient. We will compute each term for \( r = 0, 1, 2, 3, 4, 5 \).

Step 1: Term Calculation

- For \( r = 0 \): \( \frac{{}^{11}C_1}{2} = \frac{11}{2} \)
- For \( r = 1 \): \( \frac{{}^{11}C_3}{4} = \frac{165}{4} \)
- For \( r = 2 \): \( \frac{{}^{11}C_5}{6} = \frac{462}{6} = 77 \)
- For \( r = 3 \): \( \frac{{}^{11}C_7}{8} = \frac{330}{8} = 41.25 \)
- For \( r = 4 \): \( \frac{{}^{11}C_9}{10} = \frac{55}{10} = 5.5 \)
- For \( r = 5 \): \( \frac{{}^{11}C_{11}}{12} = \frac{1}{12} \).

Step 2: Summation of Terms

The sum of these terms is: \[ \frac{11}{2} + \frac{165}{4} + 77 + 41.25 + 5.5 + \frac{1}{12} \] To add these, we use a common denominator of 12: \[ \frac{66}{12} + \frac{495}{12} + \frac{924}{12} + \frac{495}{12} + \frac{66}{12} + \frac{1}{12} \] Summing the numerators: \[ \frac{66 + 495 + 924 + 495 + 66 + 1}{12} = \frac{2047}{12} \]

Step 3: Final Result

The sum evaluates to \( \frac{2047}{12} \). Given the sum is in the form \( \frac{m}{n} \), where \( m \) and \( n \) are coprime, we have \( m = 2047 \) and \( n = 12 \). The required value is \( m - n \).

\[ m - n = 2047 - 12 = 2035 \] The final answer is: \[ \boxed{2035} \]