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 the Binomial Sum Problem 

We are given the following sum:

\[ \sum_{r=0}^5 \frac{{}^{11}C_{2r+1}}{2r+2} \] The general term in the sum is \( \frac{{}^{11}C_{2r+1}}{2r+2} \), where \( {}^{11}C_k \) is the binomial coefficient. Let's calculate each term in the sum for \( r = 0, 1, 2, 3, 4, 5 \).

 

Step 1: Calculating the individual terms

- 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: Adding the terms

Now, we sum up all the terms: \[ \frac{11}{2} + \frac{165}{4} + 77 + 41.25 + 5.5 + \frac{1}{12} \] To add these fractions, we need a common denominator, which is 12. We rewrite each term with denominator 12: \[ \frac{11}{2} = \frac{66}{12}, \quad \frac{165}{4} = \frac{495}{12}, \quad 77 = \frac{924}{12}, \quad 41.25 = \frac{495}{12}, \quad 5.5 = \frac{66}{12}, \quad \frac{1}{12} = \frac{1}{12} \] Adding them up: \[ \frac{66 + 495 + 924 + 495 + 66 + 1}{12} = \frac{2047}{12} \]

Step 3: Final Answer

The sum is \( \frac{2047}{12} \). We are given that this sum is of the form \( \frac{m}{n} \), where \( m \) and \( n \) are coprime. In this case, \( m = 2047 \) and \( n = 12 \), and we are asked to find \( m - n \). \[ m - n = 2047 - 12 = 2035 \] Therefore, the final answer is: \[ \boxed{2035} \]