Question

If in the expansion of ${\left(1+x\right)}^m{\left(1-x\right)}^n$, the coefficients of $x$ and $x^2$ are $3$ and $-6$ respectively, then $m$ is

• A. $6$
• B. $9$
• C. $12$
• D. $24$

Answer 1

The correct option is C

$12$

Explanation for the correct option:

Step -1: Form the equations:

The series ${\left(1+x\right)}^m{\left(1-x\right)}^n$ can be expanded as

$\begin{array}{rcl}{\left(1+x\right)}^m{\left(1-x\right)}^n& =& \left[1+mx+\frac{m\left(m-1\right)}{2}{x}^2+...\right]\left[1-nx+\frac{n\left(n-1\right)}{2}{x}^2-...\right]\\ & =& 1+\left(m-n\right)x+\left(\frac{n\left(n-1\right)}{2}+\frac{m\left(m-1\right)}{2}-mn\right)x^2+....\end{array}$

Now it is given that the coefficient of $x$ is $3$, so $m-n=3$ and thus $n=m-3...\left(1\right)$.

Also it is given that the coefficient of $x^2$ is $-6$, so

$$\begin{gathered} \frac{n(n-1)}{2}+\frac{m(m-1)}{2}-mn=-6 \\ \Rightarrow n^2+m^2-n-m-2mn+12=0 \\ \Rightarrow (m^2+n^2-2mn)-(m+n)=-12 \\ \Rightarrow {(m-n)}^2-(m+n)=-12...\left(2\right) \end{gathered}$$

Step- 2: Find the value of $m$:

In the equation $2$, ${\left(m-n\right)}^2-\left(m+n\right)=-12$ use$m-n=3$ and solve for $m$.

$$\begin{gathered} {\left(m-n\right)}^2-\left(m+n\right)=-12 \\ \Rightarrow 3^2-\left(m+n\right)=-12\mathbf{[}\mathbf{m}\mathbf{-}\mathbf{n}\mathbf{=}\mathbf{3}\mathbf{]} \\ \Rightarrow -\left(m+m-3\right)=-12-9\left[\mathit{n}\mathbf{=}\mathit{m}\mathbf{-}\mathbf{3}\right] \\ \Rightarrow 2m-3=21 \\ \Rightarrow 2m=24 \\ \Rightarrow m=12 \end{gathered}$$

And so the value of $m$ is $12$.

Hence, the correct option is C.