Let $C_r$ denote the binomial coefficient of $x^r$ in the expansion of $(1+x{)}^{10}$. If $\alpha ,\beta \in R$. $C_1+3·2C_2+5·3C_3+\dots$ upto 10 terms $=\frac{\alpha \times 2^{11}}{2^{\beta }-1}\left({\mathrm{C}}_0+\frac{{\mathrm{C}}_1}{2}+\frac{{\mathrm{C}}_2}{3}+\dots .\right.$ upto 11 terms $)$ then the value of $\alpha +\beta$ is equal to

BONUS

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

Differentiating

$$\begin{gathered} 10(1+x{)}^9=C_1+2C_{2}x+3C_3{x}^2+\dots .+10C_{10}{x}^9 \\ \text{ replace }x\to x^2 \\ 10{\left(1+x^2\right)}^9=C_1+2C_2{x}^2+3C_3{x}^4+\dots ..+10C_{10}{x}^{18} \\ 10·x{\left(1+x^2\right)}^9=C_{1}x+2C_2{x}^3+3C_3{x}^5+\dots +10C_{10}{x}^{19} \end{gathered}$$

Differentiating

$$\begin{gathered} 10\left({\left(1+x^2\right)}^9·1+x^{·}·9{\left(1+x^2\right)}^{8}2x\right) \\ =C_{1}x+2C_2·3x^3+3·5·C_3{x}^4+\dots .+10·19C_{10}{x}^{18} \end{gathered}$$

putting x = 1

$$\begin{gathered} 10\left(2^9+18·2^8\right)=C_1+3·2·C_2+5·3·C_3+\dots +19·10·C_{10} \\ {\mathrm{C}}_1+3·2·{\mathrm{C}}_2+\dots \dots ..+19·10·{\mathrm{C}}_{10}=10·2^9·10=100·2^9 \\ {\mathrm{C}}_0+\frac{{\mathrm{C}}_1}{2}+\frac{{\mathrm{C}}_2}{3}+\dots ..+\frac{{\mathrm{C}}_9}{11}+\frac{{\mathrm{C}}_{10}}{11}=\frac{2^{11}-1}{11} \\ \uparrow \uparrow \\ {10}^{\mathrm{th}} \mathrm{term} {11}^{\mathrm{th}} \mathrm{term} \\ \text{ Now, } 100·2^9=\frac{\alpha ·2^{11}}{2^{\beta }-1}\left(\frac{2^{11}-1}{11}\right)\Rightarrow \alpha =275,\beta =11 \end{gathered}$$