If (C1 11)/(2) + (C2 11)/(3) + . . + (C9 11)/(10) = (n)/(m) with (...

Given,

$\dfrac{C1 11}{2} + \dfrac{C2 11}{3} + \dfrac{C3 11}{4} . . . . . . . . . + \dfrac{C9 11}{10} = \dfrac{m}{n}$

Now, we know that,

${\left(1 + x\right)}^{11} = C0 11 + C1 11 x + C2 11 {x}^{2} + . . . . . . . . . . C11 11 {x}^{11}$

Now integrating the above expression both side we get,

${∫}_{0}^{1}{\left(1 + x\right)}^{11} d x = {∫}_{0}^{1} C0 11 + C1 11 x + C2 11 {x}^{2} + . . . . . . . . . . C11 11 {x}^{11} d x$

$⇒ {\left[\dfrac{{\left(1 + x\right)}^{12}}{12}\right]}_{0}^{1} = {\left[C0 11 x + C1 11 \dfrac{{x}^{2}}{2} + C2 11 \dfrac{{x}^{3}}{3} + . . . . . . . . . . C11 11 \dfrac{{x}^{12}}{12}\right]}_{0}^{1}$

$⇒ \dfrac{{2}^{12} - 1}{12} = 1 + \dfrac{C1 11}{2} + \dfrac{C2 11}{3} + \dfrac{C3 11}{4} . . . . . . . . . + \dfrac{C9 11}{10} + \dfrac{11}{11} + \dfrac{1}{12}$

$⇒ \dfrac{{2}^{12} - 1}{12} - 1 - 1 - \dfrac{1}{12} = \dfrac{C1 11}{2} + \dfrac{C2 11}{3} + \dfrac{C3 11}{4} . . . . . . . . . + \dfrac{C9 11}{10}$

$⇒ \dfrac{C1 11}{2} + \dfrac{C2 11}{3} + \dfrac{C3 11}{4} . . . . . . . . . + \dfrac{C9 11}{10} = \dfrac{{2}^{12} - 2 - 24}{12}$

$⇒ \dfrac{C1 11}{2} + \dfrac{C2 11}{3} + \dfrac{C3 11}{4} . . . . . . . . . + \dfrac{C9 11}{10} = \dfrac{4096 - 26}{12}$

$⇒ \dfrac{C1 11}{2} + \dfrac{C2 11}{3} + \dfrac{C3 11}{4} . . . . . . . . . + \dfrac{C9 11}{10} = \dfrac{4070}{12}$

$⇒ \dfrac{C1 11}{2} + \dfrac{C2 11}{3} + \dfrac{C3 11}{4} . . . . . . . . . + \dfrac{C9 11}{10} = \dfrac{2035}{6}$

Hence, on comparing we get, $m + n = 2041$