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

Correct Answer: 286

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 \(286\).