Question

If in the expansion of $(1 + x)^m (1 - x)^n$, the coefficients of $x$ and $x^2$ are $3$ and - $6$ respectively, then $m$ is equal to

• A. 6
• B. 9
• C. 12
• D. 24

Answer 1

Answer: (C)

$(1+x)^m(1-x)^n= \bigg[1+mx+\frac{m(m-1)}{2}x^2+...\bigg]$
$ \, \bigg[1-nx+\frac{n(n-1)}{2}x^2+...\bigg]$
$=1+(m-n)x+\bigg[ \frac{m(m-1)}{2}+\frac{n(n-1)}{2}-mn\bigg] x^2+...$
term containing power of $x\ge 3$.
Now, $m -n = 3 $....(i)
$[\because$ coefficient of $x = 3$, given]
and $ \frac{1}{2}m(m-1)x+\frac{1}{2}n(n-1)-mn=-6$
$\Rightarrow m(m-1)+n(n-1)-2mn=-12 $
$\Rightarrow m^2-m+n^2-n-2mn=-12$
$\Rightarrow (m-n)^2-(m+n)=-12 $
$\Rightarrow m+n=9+12=21 .....(ii)$
On solving Eqs. (i) and (ii), we get $m = 12$