Question

The coefficient of the middle term in the binomial expansion in powers of $x$ of $(1+\alpha x{)}^4$ and of $(1-\alpha x{)}^6$ is the same if $\alpha$ equals :

• A. $-\frac{5}{3}$
• B. $\frac{10}{3}$
• C. $\frac{-3}{10}$
• D. $\frac{3}{5}$

Answer 1

Answer: (C)

$\frac{-3}{10}$

Given the expressions

$(1+\alpha x{)}^4$ and $(1-\alpha x{)}^6$

The coefficient of the middle of $(1+\alpha x{)}^4$ is ${}^4{C}_2{\alpha }^2$

and of $(1-\alpha x{)}^6$ is ${}^6{C}_3(-\alpha {)}^3$

Equating both coefficients

${}^4{C}_2{\alpha }^2{=}^6{C}_3(-{\alpha }^3)$

${\Rightarrow }^4{C}_2{\alpha }^2{=}^6{C}_3(-\alpha {)}^3$

${\Rightarrow }^4{C}_2{=}^6{C}_3(-\alpha )$

$\Rightarrow \frac{4!}{2!(4-2)!}=\frac{6!}{3!(6-3)!}(-\alpha )$

$\Rightarrow \frac{4!}{2!2!}=\frac{6!}{3!(3!)}(-\alpha )$

$\Rightarrow \frac{4\times 3\times 2}{2\times 2}=\frac{6\times 5\times 4\times 3\times 2\times -\alpha }{3\times 2\times 3\times 2}$

$\Rightarrow 6=20\times -\alpha$

$\Rightarrow \alpha =\frac{-6}{20}=\frac{-3}{10}$

$\therefore \alpha =\frac{-3}{10}$