Let $C_r$ denote the binomial coefficient of $x^r$ in the expansion of ${\left(1+x\right)}^{10}$. If for $\alpha ,\beta \in \mathit{R}$, $C_1+3·2C_2+5·3C_3+\dots$ upto $10$ terms $=\frac{\alpha \times 2^{11}}{2^{\beta }-1}$ ($C_0+\frac{C_1}{2}+\frac{C_2}{3}+\dots$ upto $10$ terms) then the value of $\alpha +\beta$ is equal to _____.

${\left(1+x\right)}^{10}=C_0+C_{1}x+C_2{x}^2+\dots \dots +C_{10}{x}^{10}$

Differentiating, we get

$10{\left(1+x\right)}^9=C_1+2C_{2}x+3C_3{x}^2+\dots +10C_{10}{x}^9$

Now replacing $x\to x^2$

$10{\left(1+x^2\right)}^9=C_1+2C_2{x}^2+3C_3{x}^4+\dots .+10C_{10}{x}^{18}$

Multiplying by $x$ on both the sides

$10·x{\left(1+x^2\right)}^9=C_{1}x+2C_2{x}^3+3C_3{x}^5+\dots +10C_{10}{x}^{19}$

Differentiating again, we get 

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

$=C_1+2C_2·3x^2+3·5·C_3{x}^4+\dots +10·19C_{10}{x}^{18}$

putting $x=1$

$10\left(2^9+18·2^8\right)=C_1+3·2·C_2+5·3·C_3+\dots .+19·10·C_{10}$

$C_1+3·2·C_2+\dots \dots +19·10·C_{10}$

$=10·2^9·10=100·2^9$

Now ${\left(1+x\right)}^{10}=C_0+C_{1}x+C_2{x}^2+\dots \dots +C_{10}{x}^{10}$

Integrating, we get

${\left[\frac{{\left(1+x\right)}^{11}}{11}\right]}_0^1={\left[C_{0}x+C_1\frac{x^2}{2}+\frac{C_2{x}^3}{3}+\dots \dots +\frac{C_{10}{x}^{11}}{11}\right]}_0^1$

$C_0+\frac{C_1}{2}+\frac{C_2}{3}+\dots .+\frac{C_9}{10}+\frac{C_{10}}{11}=\frac{2^{11}-1}{11}$

$C_0+\frac{C_1}{2}+\frac{C_2}{3}+\dots .+\frac{C_9}{10}=\frac{2^{11}-2}{11}$

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

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