Question

Let $C_r$ denote the binomial coefficient of $x^r$ in the expansion of $(1+x{)}^{10}$. If $\alpha ,\beta \in R$. $C_1+3\cdot 2C_2+5\cdot 3C_3+\dots$ 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

Answer 1

Answer: 286

$(1+x{)}^{10}=C_0+C_{1}x+C_2{x}^2+\dots \dots +C_{10}{x}^{10}$
Differentiating
$10(1+x{)}^9=C_1+2C_{2}x+3C_3{x}^2+\dots .+10C_{10}{x}^9$
replace $x\to x^2$
$10{\left(1+x^2\right)}^9=C_1+2C_2{x}^2+3C_3{x}^4+\dots +10C_{10}{x}^{18}$
$10\cdot x{\left(1+x^2\right)}^9=C_{1}x+2C_2{x}^3+3C_3{x}^5+\dots +10C_{10}{x}^{19}$
Differentiating
$10\left({\left(1+x^2\right)}^9\cdot 1+x\cdot 9{\left(1+x^2\right)}^{8}2x\right)$
$=C_{1}x+2C_2\cdot 3x^3+3\cdot 5\cdot C_3{x}^4+\dots .+10\cdot 19C_{10}{x}^{18}$
putting $x=1$
$10\left(2^9+18\cdot 2^8\right)$
$=C_1+3\cdot 2\cdot C_2+5\cdot 3\cdot C_3+\dots +19\cdot 10\cdot C_{10}$
$C_1+3\cdot 2\cdot C_2+\dots \dots +19\cdot 10\cdot C_{10}$
$=10\cdot 2^9\cdot 10=100\cdot 2^9$

$C_0+\frac{C_1}{2}+\frac{C_2}{3}+\dots .+\frac{C_9}{11}=\frac{2^{11}-2}{11}$
Now, $100\cdot 2^9=\frac{\alpha \cdot 2^{11}}{2^{\beta }-1}\left(\frac{2^{11}-2}{11}\right)$
Eqn. of form $y=k\left(2^x-1\right)$.
It has infinite solutions even if we take $x,y\in N$.