Question

Let $C_r$ denote the binomial coefficient of $x^r$in the expansion of $(1 + x)^{10}$. If for $α, β∈R$, $C_1 \+ 3⋅2 C_2 \+ 5⋅3 C_3 \+ …$ upto 10 terms =$\frac {α×211}{2^β−1}(C_0+\frac {C_1}{2}+\frac {C_2}{3}…..$upto 10 terms) then the value of $α + β$ is equal to _______.

Answer 1

Given that $C_1 \+ 2⋅3C_2 \+ 5⋅3C_3 \+ … 10$ terms
$= \frac {α⋅2^{11}}{2^β−1}(C_1+\frac {C_2}{2}+…...)$

$\displaystyle\sum_{r=1}^{10} r(2r−1)C_r= \frac {α⋅2^{11}}{2^β−1} (\displaystyle\sum_{r=1}^{10}\frac {C_r}{r})$

Using $C_1 \+ 2C_2 \+ …. + nC_n = n.2^{n – 1}$
$1^2C_1 \+ 2^2C_2 \+ … + n^2C_n = n.^{2n – 1} + n(n – 1)2^{n – 2}$
and,
$C_0+\frac {C_1}{2}+…...\frac {C_n}{n+1}=\frac {2^{n+1}−1}{n+1}$
we get,
$2(10.2^9 \+ 10.9.2^8) – 10.2^9$
$=\frac {α⋅2^{11}}{2^β−1}\frac {(2^{11}−1)}{11}$
On comparing both side,
2^{11}.25=$$\frac {α⋅2^{11}}{2^β−1}\frac {(2^{11}−1)}{11}
$⇒ α = 25 × 11 = 275$
$β = 11$
$⇒ α + β = 286$

So, the answer is $286$.