If C 1 11 2 + C 2 11 3 + ? + C 9 11 10 = n m with gcd ( n , m ) = 1 , then n + m is equal to _______ . [2024]

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Solution

Q.
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 _______ . [2024]

Ans.
(2041)

$\frac{{}^{11}{C_{1}}}{2} + \frac{{}^{11}{C_{2}}}{3} + \dots + \frac{{}^{11}{C_{9}}}{10}$

$= ({}^{11}{C_{0}} + \frac{{}^{11}{C_{1}}}{2} + \frac{{}^{11}{C_{2}}}{3} + \dots + \frac{{}^{11}{C_{9}}}{10} + \frac{{}^{11}{C_{10}}}{11} + \frac{{}^{11}{C_{11}}}{12})$ $- 1 - 1 - \frac{1}{12}$

$= \frac{2^{11 + 1} - 1}{12} - 2 - \frac{1}{12} = \frac{2^{12} - 26}{12} = \frac{4070}{12} = \frac{2035}{6}$

$∴$ $n + m = 2035 + 6 = 2041$

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