Question

Let $C _{ r }$ denote the binomial coefficient of $x ^{ r }$ in the expansion of $(1+x)^{10}$. If $\alpha, \beta \in R$. $C_{1}+3 \cdot 2 C_{2}+5 \cdot 3 C_{3}+\ldots$ upto $10$ terms $=\frac{\alpha \times 2^{11}}{2^{\beta}-1}\left( C _{0}+\frac{ C _{1}}{2}+\frac{ C _{2}}{3}+\ldots .\right.$ upto $10$ terms $)$ then the value of $\alpha+\beta$ is equal to

Answer 1

$(1+x)^{10}=C_{0}+C_{1} x+C_{2} x^{2}+\ldots \ldots+C_{10} x^{10}$
Differentiating
$10(1+x)^{9}=C_{1}+2 C_{2} x+3 C_{3} x^{2}+\ldots .+10 C_{10} x^{9}$
replace $x \rightarrow x^{2}$
$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} $
$10 \cdot x\left(1+x^{2}\right)^{9}=C_{1} x+2 C_{2} x^{3}+3 C_{3} x^{5}+\ldots+10 C_{10} x^{19}$
Differentiating
$10\left(\left(1+x^{2}\right)^{9} \cdot 1+x \cdot 9\left(1+x^{2}\right)^{8} 2 x\right) $
$=C_{1} x+2 C_{2} \cdot 3 x^{3}+3 \cdot 5 \cdot C_{3} x^{4}+\ldots .+10 \cdot 19 C_{10} x^{18}$
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}$
$=10 \cdot 2^{9} \cdot 10=100 \cdot 2^{9}$

$C _{0}+\frac{ C _{1}}{2}+\frac{ C _{2}}{3}+\ldots .+\frac{ C _{9}}{11}=\frac{2^{11}-2}{11}$
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 $x, y \in N$.