Let C _ r denote the binomial coefficient of x ^ r in the expansion...

${\left(1 + x\right)}^{10} = {C}_{0} + {C}_{1}x + {C}_{2}{x}^{2} + \ldots \ldots + {C}_{10}{x}^{10}$

Differentiating, we get

$10{\left(1 + x\right)}^{9} = {C}_{1} + 2{C}_{2}x + 3{C}_{3}{x}^{2} + \ldots + 10{C}_{10}{x}^{9}$

Now replacing $x \to {x}^{2}$

$10{\left(1 + {x}^{2}\right)}^{9} = {C}_{1} + 2{C}_{2}{x}^{2} + 3{C}_{3}{x}^{4} + \ldots . + 10{C}_{10}{x}^{18}$

Multiplying by $x$ on both the sides

$10 \cdot x{\left(1 + {x}^{2}\right)}^{9} = {C}_{1}x + 2{C}_{2}{x}^{3} + 3{C}_{3}{x}^{5} + \ldots + 10{C}_{10}{x}^{19}$

Differentiating again, we get

$10\left({\left(1 + {x}^{2}\right)}^{9} \cdot 1 + 9{\left(1 + {x}^{2}\right)}^{8}2x\right)$

$= {C}_{1} + 2{C}_{2} \cdot 3{x}^{2} + 3 \cdot 5 \cdot {C}_{3}{x}^{4} + \ldots + 10 \cdot 19{C}_{10}{x}^{18}$

putting $x = 1$

$10\left({2}^{9} + 18 \cdot {2}^{8}\right) = {C}_{1} + 3 \cdot 2 \cdot {C}_{2} + 5 \cdot 3 \cdot {C}_{3} + \ldots . + 19 \cdot 10 \cdot {C}_{10}$

${C}_{1} + 3 \cdot 2 \cdot {C}_{2} + \ldots \ldots + 19 \cdot 10 \cdot {C}_{10}$

$= 10 \cdot {2}^{9} \cdot 10 = 100 \cdot {2}^{9}$

Now ${\left(1 + x\right)}^{10} = {C}_{0} + {C}_{1}x + {C}_{2}{x}^{2} + \ldots \ldots + {C}_{10}{x}^{10}$

Integrating, we get

${\left[\dfrac{{\left(1 + x\right)}^{11}}{11}\right]}_{0}^{1} = {\left[{C}_{0}x + {C}_{1}\dfrac{{x}^{2}}{2} + \dfrac{{C}_{2}{x}^{3}}{3} + \ldots \ldots + \dfrac{{C}_{10}{x}^{11}}{11}\right]}_{0}^{1}$

${C}_{0} + \dfrac{{C}_{1}}{2} + \dfrac{{C}_{2}}{3} + \ldots . + \dfrac{{C}_{9}}{10} + \dfrac{{C}_{10}}{11} = \dfrac{{2}^{11} - 1}{11}$

${C}_{0} + \dfrac{{C}_{1}}{2} + \dfrac{{C}_{2}}{3} + \ldots . + \dfrac{{C}_{9}}{10} = \dfrac{{2}^{11} - 2}{11}$

Now, $100 \cdot {2}^{9} = \dfrac{\alpha \cdot {2}^{11}}{{2}^{\beta} - 1}\left(\dfrac{{2}^{11} - 2}{11}\right)$

$⇒ \alpha = 275, \beta = 11 ⇒ \alpha + \beta = 286$