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:
Show Hint
Correct Answer: 2035
Approach Solution - 1
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: \[ \boxed{2035} \]
Approach Solution -2
Given expression: \[ S = \sum_{r=0}^{5} \frac{{}^{11}C_{2r+1}}{2r+2}. \] Let’s manipulate it using binomial expansion properties.
Consider expansion of \((1+x)^{11}\): \[ (1+x)^{11} = \sum_{k=0}^{11} {}^{11}C_k x^k. \] We separate odd and even terms using: \[ (1+x)^{11} + (1-x)^{11} = 2\sum_{\text{even }k} {}^{11}C_k x^k, \] \[ (1+x)^{11} - (1-x)^{11} = 2\sum_{\text{odd }k} {}^{11}C_k x^k. \] Hence, \[ \sum_{\text{odd }k} {}^{11}C_k x^k = \frac{(1+x)^{11} - (1-x)^{11}}{2}. \]
Now, in our sum, \(k = 2r + 1\). So, \[ S = \sum_{r=0}^{5} \frac{{}^{11}C_{2r+1}}{2r+2} = \sum_{\text{odd }k} \frac{{}^{11}C_k}{k+1}. \] This can be written as: \[ S = \int_0^1 \sum_{\text{odd }k} {}^{11}C_k x^k\,dx = \int_0^1 \frac{(1+x)^{11} - (1-x)^{11}}{2} \, dx. \]
Compute the integral: \[ S = \frac{1}{2} \int_0^1 \big((1+x)^{11} - (1-x)^{11}\big) dx. \] Integrate each term separately: \[ \int (1+x)^{11}dx = \frac{(1+x)^{12}}{12}, \quad \int (1-x)^{11}dx = -\frac{(1-x)^{12}}{12}. \]
Therefore, \[ S = \frac{1}{2} \left[ \frac{(1+x)^{12} + (1-x)^{12}}{12} \right]_0^1. \] Simplify: \[ S = \frac{1}{24} \left[ (1+1)^{12} + (1-1)^{12} - \big( (1+0)^{12} + (1-0)^{12} \big) \right]. \] Compute values: \[ S = \frac{1}{24} \big[ (2^{12}+0) - (1+1) \big] = \frac{1}{24} (4096 - 2) = \frac{4094}{24}. \]
Simplify the fraction: \[ \frac{4094}{24} = \frac{2047}{12}. \] Since \(\gcd(2047,12)=1\), we have \(m=2047, n=12\).
Therefore, \[ m-n = 2047 - 12 = \boxed{2035}. \]
Answer
\(\boxed{2035}\)
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