Q. If in the expansion of $({1 + x)}^{m}({1 - x)}^{n},$ the coefficient of $x$ and $x^{2}$ are 3 and $- 6$ respectively, then $m$ is

A) 6
B) 9
C) 12
D) 24
Explanation

(c)
Given , $(1 + x)^{m}({1 - x)}^{n}$
$= \left( 1 + mx + m\frac{(m - 1)}{2!}x^{2} + \ldots \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( 1 - nx + \frac{n(n - 1)}{2!}x^{2} - \ldots \right)$
$= 1 + (m - n)x + \left\lbrack \frac{n^{2} - n}{2} - mn + \frac{m^{2} - m}{2} \right\rbrack x^{2} + \ldots$
Also, given $m - n = 3\ \Longrightarrow n = m - 3$
$and\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{n^{2} - n}{2} - mn + \frac{m^{2} - m}{2} = - 6$
$\Longrightarrow \ \ \ \ \ \ \ \frac{(m - 3)(m - 4)}{2} - m(m - 3) + \frac{m^{2} - m}{2} - 6$
$\Longrightarrow \ m^{2} - 7m + 12 - 2m^{2} + 6m + m^{2} - m + 12 = 0$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - 2m + 24 = 0\ \Longrightarrow m = 12\$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.