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)

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