1. Key Concept: Integration of Binomial Expansion

The problem requires us to evaluate a sum involving binomial coefficients of the form $\frac{{}^nC_r}{r+k}$. Such sums can often be found by integrating the binomial expansion of $(1+x)^n$ or related functions. The Binomial Theorem states that for any positive integer $n$: $(1+x)^n = {^nC_0} + {^nC_1}x + {^nC_2}x^2 + \cdots + {^nC_n}x^n = \sum_{r=0}^n {^nC_r}x^r$ Integrating this expansion term by term with respect to $x$ will introduce terms of the form $\frac{{}^nC_r}{r+1}x^{r+1}$. By choosing appropriate limits of integration, we can isolate specific sums. In this particular problem, we have a sum of odd-indexed coefficients divided by an even number, which suggests that combining integrals over different ranges will be beneficial.

2. Binomial Expansion and General Integral Form

For the given problem, $n=11$. So, the binomial expansion of $(1+x)^{11}$ is: $(1+x)^{11} = {^{11}C_0} + {^{11}C_1}x + {^{11}C_2}x^2 + \cdots + {^{11}C_{11}}x^{11}$ To get terms involving division by $r+k$, we integrate. The general form of the integral of $(1+x)^{11}$ is: $\int (1+x)^{11} dx = \frac{(1+x)^{12}}{12} + C$ And integrating the series term by term: $\int \left( {^{11}C_0} + {^{11}C_1}x + {^{11}C_2}x^2 + \cdots + {^{11}C_{11}}x^{11} \right) dx = {^{11}C_0}x + \frac{{^{11}C_1}x^2}{2} + \frac{{^{11}C_2}x^3}{3} + \cdots + \frac{{^{11}C_{11}}x^{12}}{12} + C'$ We will now use definite integrals with specific limits to generate the required sum.

3. Integration from $x=0$ to $x=1$

We integrate both sides of the binomial expansion from $x=0$ to $x=1$. Why: Integrating from $0$ to $1$ simplifies $x^k$ to $1/(k+1)$ and results in a sum of all coefficients divided by their respective power plus one, which is a common form for these types of sums.

$\int_0^1 (1+x)^{11} dx = \int_0^1 \left( {^{11}C_0} + {^{11}C_1}x + {^{11}C_2}x^2 + \cdots + {^{11}C_{11}}x^{11} \right) dx$ Evaluating the left-hand side (LHS): $\text{LHS} = \left[ \frac{(1+x)^{12}}{12} \right]_0^1 = \frac{(1+1)^{12}}{12} - \frac{(1+0)^{12}}{12} = \frac{2^{12}-1^{12}}{12} = \frac{2^{12}-1}{12}$ Evaluating the right-hand side (RHS) by integrating term by term: $\text{RHS} = \left[ {^{11}C_0}x + \frac{{^{11}C_1}x^2}{2} + \frac{{^{11}C_2}x^3}{3} + \cdots + \frac{{^{11}C_{11}}x^{12}}{12} \right]_0^1$ $\text{RHS} = \left( {^{11}C_0}(1) + \frac{{^{11}C_1}(1)^2}{2} + \frac{{^{11}C_2}(1)^3}{3} + \cdots + \frac{{^{11}C_{11}}(1)^{12}}{12} \right) - (0)$ $\text{RHS} = {^{11}C_0} + \frac{{^{11}C_1}}{2} + \frac{{^{11}C_2}}{3} + \cdots + \frac{{^{11}C_{11}}}{12} \quad \cdots (1)$ So, we have: ${^{11}C_0} + \frac{{^{11}C_1}}{2} + \frac{{^{11}C_2}}{3} + \cdots + \frac{{^{11}C_{11}}}{12} = \frac{2^{12}-1}{12}$

4. Integration from $x=-1$ to $x=0$

Next, we integrate both sides of the binomial expansion from $x=-1$ to $x=0$. Why: Integrating with $-1$ as a limit introduces alternating signs for terms with odd powers of $x$, which will be crucial for isolating the odd-indexed coefficients when combined with the previous result.

$\int_{-1}^0 (1+x)^{11} dx = \int_{-1}^0 \left( {^{11}C_0} + {^{11}C_1}x + {^{11}C_2}x^2 + \cdots + {^{11}C_{11}}x^{11} \right) dx$ Evaluating the left-hand side (LHS): $\text{LHS} = \left[ \frac{(1+x)^{12}}{12} \right]_{-1}^0 = \frac{(1+0)^{12}}{12} - \frac{(1-1)^{12}}{12} = \frac{1^{12}-0^{12}}{12} = \frac{1}{12}$ Evaluating the right-hand side (RHS) by integrating term by term: $\text{RHS} = \left[ {^{11}C_0}x + \frac{{^{11}C_1}x^2}{2} + \frac{{^{11}C_2}x^3}{3} + \cdots + \frac{{^{11}C_{11}}x^{12}}{12} \right]_{-1}^0$ $\text{RHS} = (0) - \left( {^{11}C_0}(-1) + \frac{{^{11}C_1}(-1)^2}{2} + \frac{{^{11}C_2}(-1)^3}{3} + \cdots + \frac{{^{11}C_{11}}(-1)^{12}}{12} \right)$ $\text{RHS} = - \left( -{^{11}C_0} + \frac{{^{11}C_1}}{2} - \frac{{^{11}C_2}}{3} + \frac{{^{11}C_3}}{4} - \cdots + \frac{{^{11}C_{11}}}{12} \right)$ $\text{RHS} = {^{11}C_0} - \frac{{^{11}C_1}}{2} + \frac{{^{11}C_2}}{3} - \frac{{^{11}C_3}}{4} + \cdots - \frac{{^{11}C_{11}}}{12} \quad \cdots (2)$ So, we have: ${^{11}C_0} - \frac{{^{11}C_1}}{2} + \frac{{^{11}C_2}}{3} - \frac{{^{11}C_3}}{4} + \cdots - \frac{{^{11}C_{11}}}{12} = \frac{1}{12}$ Tip: Pay close attention to the signs when evaluating at the lower limit and subtracting, especially for alternating series. An easier way to think about the alternating series is to consider the original expansion of $(1-x)^{11}$ which would result in ${^{11}C_0} - {^{11}C_1}x + {^{11}C_2}x^2 - \dots$, and then integrate that from $0$ to $1$. However, the method used here is also valid.

5. Combining the Results

Now, we subtract equation (2) from equation (1). Why: Subtracting these two equations will cause all terms with even-indexed coefficients ($C_0, C_2, C_4, \ldots$) to cancel out, while all terms with odd-indexed coefficients ($C_1, C_3, C_5, \ldots$) will be added together (doubled). This isolates the sum we need.

$(1) - (2):$ $\left( \frac{2^{12}-1}{12} \right) - \left( \frac{1}{12} \right) = \left( {^{11}C_0} + \frac{{^{11}C_1}}{2} + \frac{{^{11}C_2}}{3} + \cdots + \frac{{^{11}C_{11}}}{12} \right) - \left( {^{11}C_0} - \frac{{^{11}C_1}}{2} + \frac{{^{11}C_2}}{3} - \cdots - \frac{{^{11}C_{11}}}{12} \right)$ Simplifying the LHS: $\text{LHS} = \frac{2^{12}-1-1}{12} = \frac{2^{12}-2}{12}$ Simplifying the RHS: $\text{RHS} = 2 \left( \frac{{^{11}C_1}}{2} + \frac{{^{11}C_3}}{4} + \frac{{^{11}C_5}}{6} + \frac{{^{11}C_7}}{8} + \frac{{^{11}C_9}}{10} + \frac{{^{11}C_{11}}}{12} \right)$ The sum on the RHS matches the form $\sum_{r=0}^5 \frac{{ }^{11} C_{2 r+1}}{2 r+2}$. So, we have: $\frac{2^{12}-2}{12} = 2 \sum_{r=0}^5 \frac{{ }^{11} C_{2 r+1}}{2 r+2}$ Divide by 2 to find the required sum: $\sum_{r=0}^5 \frac{{ }^{11} C_{2 r+1}}{2 r+2} = \frac{2^{12}-2}{2 \times 12} = \frac{2(2^{11}-1)}{24} = \frac{2^{11}-1}{12}$

6. Final Calculation and GCD Check

Now we calculate the numerical value of the sum: We know $2^{11} = 2048$. $\sum_{r=0}^5 \frac{{ }^{11} C_{2 r+1}}{2 r+2} = \frac{2048-1}{12} = \frac{2047}{12}$ The problem states that the sum is $\frac{m}{n}$ where $\operatorname{gcd}(m, n)=1$. Here, $m = 2047$ and $n = 12$. We need to check if $m$ and $n$ are coprime. The prime factors of $n=12$ are $2$ and $3$. $2047$ is not divisible by $2$ (it's an odd number). Sum of digits of $2047 = 2+0+4+7 = 13$, which is not divisible by $3$. Therefore, $\operatorname{gcd}(2047, 12)=1$. Finally, we need to find $m-n$: $m-n = 2047 - 12 = 2035$

7. Tips for Success

• Recognize the pattern: When you see binomial coefficients divided by $r+k$, think about integrating the binomial expansion.
• Choose limits wisely: The limits of integration ($0$ to $1$, $-1$ to $0$, etc.) are chosen to simplify the resulting sum and isolate the desired terms.
• Careful with signs: When evaluating definite integrals, especially with negative lower limits, be meticulous with the signs. An error in one sign can completely change the result.
• Check GCD: Always perform the $\operatorname{gcd}$ check to ensure the fraction is in its simplest form before calculating $m-n$.

8. Key Takeaway

This problem demonstrates a powerful technique for evaluating sums involving binomial coefficients divided by linear terms using integral calculus. By integrating the binomial expansion $(1+x)^n$ over suitable intervals and combining the resulting series, we can selectively isolate and sum odd or even indexed binomial coefficients. This method is a versatile tool in combinatorial problems.