Key Concept: The Identity for Binomial Coefficients with Denominators

This problem requires the use of a specific identity for binomial coefficients that simplifies expressions involving $\frac{{}^n C_r}{r+1}$. The key identity is: $\frac{{}^n C_r}{r+1} = \frac{1}{n+1} \cdot {}^{n+1} C_{r+1}$

Explanation of the Identity: This identity is incredibly useful because it transforms a term that is difficult to sum directly into a standard binomial coefficient, making it suitable for summation using the binomial theorem.

To prove this identity, we can write out the definitions: Left-hand side (LHS): $\frac{{}^n C_r}{r+1} = \frac{1}{r+1} \cdot \frac{n!}{r!(n-r)!} = \frac{n!}{(r+1)r!(n-r)!} = \frac{n!}{(r+1)!(n-r)!}$ Right-hand side (RHS): $\frac{1}{n+1} \cdot {}^{n+1} C_{r+1} = \frac{1}{n+1} \cdot \frac{(n+1)!}{(r+1)!((n+1)-(r+1))!} = \frac{1}{n+1} \cdot \frac{(n+1)!}{(r+1)!(n-r)!}$ $= \frac{n!}{(r+1)!(n-r)!}$ Since LHS = RHS, the identity is proven.

Step-by-Step Solution

1. Apply the Identity to the General Term: The given sum is $\sum_{r=1}^9 \frac{{}^{11} C_r}{r+1}$. Comparing the general term $\frac{{}^{11} C_r}{r+1}$ with the identity $\frac{{}^n C_r}{r+1}$, we have $n=11$. Applying the identity, we transform the general term: $\frac{{}^{11} C_r}{r+1} = \frac{1}{11+1} \cdot {}^{11+1} C_{r+1} = \frac{1}{12} \cdot {}^{12} C_{r+1}$ Why this step? This substitution converts the complex fraction into a simpler binomial coefficient form, which is easier to sum.

2. Rewrite the Summation: Now, substitute this transformed term back into the summation: $\sum_{r=1}^9 \frac{{}^{11} C_r}{r+1} = \sum_{r=1}^9 \frac{1}{12} \cdot {}^{12} C_{r+1}$ We can factor out the constant $\frac{1}{12}$: $= \frac{1}{12} \sum_{r=1}^9 {}^{12} C_{r+1}$ Next, we adjust the limits of the summation. Let $k = r+1$. When $r=1$, $k=1+1=2$. When $r=9$, $k=9+1=10$. So the sum becomes: $= \frac{1}{12} \sum_{k=2}^{10} {}^{12} C_k$ Why this step? Adjusting the limits and changing the summation variable ($r$ to $k$) is crucial to correctly represent the terms being summed after the transformation. This makes the sum directly comparable to parts of the standard binomial expansion.

3. Recognize the Binomial Expansion: We know the binomial expansion formula for $(1+x)^N$: $(1+x)^N = {^N C_0} + {^N C_1}x + {^N C_2}x^2 + \ldots + {^N C_N}x^N$ When $x=1$, this simplifies to the sum of all binomial coefficients: $\sum_{k=0}^N {^N C_k} = {^N C_0} + {^N C_1} + \ldots + {^N C_N} = (1+1)^N = 2^N$ For our problem, $N=12$, so the complete sum of binomial coefficients is $\sum_{k=0}^{12} {}^{12} C_k = 2^{12}$.

4. Complete the Series: Our current sum is $\sum_{k=2}^{10} {}^{12} C_k$. This sum does not include all terms from $k=0$ to $k=12$. To use the property $\sum_{k=0}^{12} {}^{12} C_k = 2^{12}$, we need to add and subtract the missing terms. The full sum is ${}^{12} C_0 + {}^{12} C_1 + {}^{12} C_2 + \ldots + {}^{12} C_{10} + {}^{12} C_{11} + {}^{12} C_{12}$. The terms missing from our sum $\sum_{k=2}^{10} {}^{12} C_k$ are ${}^{12} C_0$, ${}^{12} C_1$, ${}^{12} C_{11}$, and ${}^{12} C_{12}$. Therefore, we can write: $\sum_{k=2}^{10} {}^{12} C_k = \left( \sum_{k=0}^{12} {}^{12} C_k \right) - {}^{12} C_0 - {}^{12} C_1 - {}^{12} C_{11} - {}^{12} C_{12}$ Now, we calculate the values of these individual terms:

• ${}^{12} C_0 = 1$ (There is 1 way to choose 0 items from 12)
• ${}^{12} C_1 = 12$ (There are 12 ways to choose 1 item from 12)
• ${}^{12} C_{11} = {}^{12} C_{12-11} = {}^{12} C_1 = 12$ (Using the property ${}^n C_r = {}^n C_{n-r}$)
• ${}^{12} C_{12} = 1$ (There is 1 way to choose 12 items from 12)

Substituting these values and $2^{12}$: $\sum_{k=2}^{10} {}^{12} C_k = 2^{12} - 1 - 12 - 12 - 1$ $= 2^{12} - (1+12+12+1) = 2^{12} - 26$ Why this step? By completing the series, we leverage a well-known and easily calculable sum ($2^N$), allowing us to find the value of our partial sum by subtraction.

5. Simplify the Expression: Now, substitute this back into the overall expression: $\frac{1}{12} (2^{12} - 26)$ First, calculate $2^{12}$: $2^{10} = 1024$ $2^{12} = 2^{10} \times 2^2 = 1024 \times 4 = 4096$ So, the expression becomes: $\frac{1}{12} (4096 - 26) = \frac{1}{12} (4070)$ Simplify the fraction by dividing both numerator and denominator by their greatest common divisor. Both are divisible by 2: $\frac{4070 \div 2}{12 \div 2} = \frac{2035}{6}$ Why this step? Performing the arithmetic precisely and simplifying the fraction brings us to the required $\frac{n}{m}$ form.

6. Verify the $\operatorname{gcd}(n,m)=1$ Condition: The problem states that $\frac{n}{m}$ must be in simplest form, meaning $\operatorname{gcd}(n,m)=1$. Here, $n=2035$ and $m=6$. Let's find the prime factors of $n$ and $m$: $m = 6 = 2 \times 3$ $n = 2035$. It ends in 5, so it's divisible by 5: $2035 = 5 \times 407$ To check for divisibility by small primes for 407: Not divisible by 2, 3 (sum of digits $4+0+7=11$), 5. Try 7: $407 \div 7 \approx 58$ (not exact) Try 11: $407 = 11 \times 37$ So, $n = 5 \times 11 \times 37$. The prime factors of $n$ are 5, 11, 37. The prime factors of $m$ are 2, 3. Since there are no common prime factors, $\operatorname{gcd}(2035, 6) = 1$. The condition is satisfied. Why this step? This is an explicit requirement of the problem statement to ensure the fraction is unique and in its simplest form. Failing to verify this could lead to an incorrect $n+m$ if the fraction was not fully simplified.

7. Calculate $n+m$: Finally, we need to find $n+m$: $n+m = 2035 + 6 = 2041$

Tips and Common Mistakes:

• Memorize or Derive the Identity: The identity $\frac{{}^n C_r}{r+1} = \frac{1}{n+1} {}^{n+1} C_{r+1}$ is fundamental for problems of this type. Practice deriving it if you find it hard to remember.
• Correctly Adjust Limits: When changing the summation variable (e.g., $r$ to $k=r+1$), always double-check the new lower and upper limits of the sum. A common mistake is to forget to adjust these.
• Be Careful with Missing Terms: When completing a series to use a known sum (like $2^N$), ensure you correctly identify all the terms that are missing from the full series and subtract them. Pay attention to the starting and ending indices.
• Simplify the Fraction: Always simplify the resulting fraction $\frac{n}{m}$ to its lowest terms before identifying $n$ and $m$ to ensure the $\operatorname{gcd}(n,m)=1$ condition is met.

Summary/Key Takeaway: This problem elegantly demonstrates how a specific identity can transform a seemingly complex sum of binomial coefficients into a form that can be solved by leveraging the well-known sum of all binomial coefficients ($2^N$). The key steps involve applying the identity, carefully adjusting summation limits, and then using subtraction to isolate the desired portion of a complete binomial expansion. Always remember to simplify the fraction to its lowest terms as required by the problem.