If frac{{}^{11}C_1}{2} + frac{{}^{11}C_2}{3} | vedclass

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(A) We use the identity $\frac{{}^nC_r}{r+1} = \frac{{}^{n+1}C_{r+1}}{n+1}$.Given the sum $S = \sum_{r=1}^9 \frac{{}^{11}C_r}{r+1}$.Applying the ident

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$2041$

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(A) We use the identity $\frac{{}^nC_r}{r+1} = \frac{{}^{n+1}C_{r+1}}{n+1}$.Given the sum $S = \sum_{r=1}^9 \frac{{}^{11}C_r}{r+1}$.Applying the identity,$S = \sum_{r=1}^9 \frac{{}^{12}C_{r+1}}{12} = \frac{1}{12} \sum_{k=2}^{10} {}^{12}C_k$.We know that $\sum_{k=0}^{12} {}^{12}C_k = 2^{12} = 4096$.Therefore,$\sum_{k=2}^{10} {}^{12}C_k = 2^{12} - ({}^{12}C_0 + {}^{12}C_1 + {}^{12}C_{11} + {}^{12}C_{12})$.$= 4096 - (1 + 12 + 12 + 1) = 4096 - 26 = 4070$.Thus,$S = \frac{4070}{12} = \frac{2035}{6}$.Since $\gcd(2035, 6) = 1$,we have $n = 2035$ and $m = 6$.Therefore,$n + m = 2035 + 6 = 2041$.

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

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