Question

If $\frac{{^{11}C_1}}{2} + \frac{{^{11}C_2}}{3} + \ldots + \frac{{^{11}C_9}}{10} = \frac{n}{m}$ with $\gcd(n, m) = 1$, then $n + m$ is equal to:

Answer 1

Step 1: Rewrite the Sum in Terms of a Series

The sum can be expressed as:

$\sum_{r=1}^{9} \left( ^{11}C_r \right) \cdot \left( ^{11}C_{r+1} \right)$

Step 2: Simplify the Series Using Combinatorial Identities

This can be simplified by recognizing a pattern and using properties of binomial coefficients:

$= \frac{1}{12} \sum_{r=1}^{9} \left( ^{12}C_{r+1} \right)$

Further simplifying, we get:

$= \frac{1}{12} \left[2^{12} - 26\right] = \frac{2035}{6}$

Step 3: Determine $n$ and $m$

From the simplified result, $n = 2035$ and $m = 6$, with $\gcd(n, m) = 1$.

Step 4: Calculate $n + m$

$n + m = 2035 + 6 = 2041$

So, the correct answer is: 2041