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

The integral \( \int_0^1 (1 + x)^{11} \, dx \) expands as: \[ \int_0^1 (1 + x)^{11} \, dx = C_0 + C_1 x^2 + C_2 x^3 + \dots \] Evaluating this, we get: \[ \frac{2^{12} - 1}{12} = C_0 + C_1 + C_2 + C_3 + \dots \] Similarly, the integral from \( -1 \) to \( 0 \) is: \[ \int_{-1}^0 (1 + x)^{11} \, dx = C_0 - C_1 + C_2 - C_3 + \dots \] From this, we can calculate: \[ \frac{2^{12} - 2}{12} = 2 \left( C_1 + C_3 + C_5 + \dots \right) \] Thus: \[ C_1 + C_3 + C_5 + \dots = \frac{2^{11} - 1}{12} = \frac{2047}{12} \] Hence, \( m - n = 2035 \).