If $\dfrac{{}^{11}C_1}{2} + \dfrac{{}^{11}C_2}{3} + \ldots + \dfrac{{}^{11}C_9}{10} = \dfrac{n}{m}$ with $\gcd(n,m)=1$, then $n+m$ is equal to
Step-by-Step Solution
Key Concept: Use the identity $\frac{{}^{11}C_r}{r+1} = \frac{1}{12}{}^{12}C_{r+1}$. Convert the sum to $\frac{1}{12}\sum_{r=1}^{9}{}^{12}C_{r+1}$, then use $\sum_{k=0}^{12}{}^{12}C_k = 2^{12}$ with boundary terms subtracted.
$\sum_{r=1}^{9}\frac{{}^{11}C_r}{r+1}=\frac{1}{12}\sum_{r=1}^{9}{}^{12}C_{r+1}=\frac{1}{12}\sum_{j=2}^{10}{}^{12}C_j$.
$\sum_{j=0}^{12}{}^{12}C_j=2^{12}=4096$. Subtract: ${}^{12}C_0+{}^{12}C_1+{}^{12}C_{11}+{}^{12}C_{12}=1+12+12+1=26$.
So sum $=\frac{1}{12}(4096-26)=\frac{4070}{12}=\frac{2035}{6}$.
$n=2035, m=6$, $\gcd(2035,6)=1$. $n+m=2041$.
Correct Answer: 2041