Question

The coefficient of the middle term in the binomial expansion in power 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

The correct option is C $\frac{-3}{10}$

$Middletermnis=mn{\left(1+x\right)}^n{T}_{r+1}{=}^n{C}_r{\left(1\right)}^{n-r}{x}^r$

$\begin{array}{c}\\ \frac{T_n}{2}+1{\left(1+\alpha x\right)}^4\\ T_3=T_{2+1}{=}^4{C}_2{\left(1\right)}^2{\left(\alpha x\right)}^2\\ {\left(i-\alpha x\right)}^6\\ T_4=T_{3+1}{=}^6{C}_3{\left(-\alpha x\right)}^3\\ T_5=T_4\\ {}^4{C}_2{\left(\alpha x\right)}^2{=}^6{C}_3{\left(-\alpha x\right)}^3\\ \frac{41}{2121}{\alpha }^2=\frac{61}{3131}{\left(-\alpha \right)}^3\\ \frac{4\times 3}{2}{\alpha }^2=\frac{6\times 1\times 4}{3\times 2}{\left(-\alpha \right)}^3\\ \frac{6}{20}=-\alpha \\ \end{array}$

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

Hence, Option $C$ is the correct answer.