Question

In the expansion $(1+x{)}^m(1-x{)}^n$ 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

Answer: (C)

$12$

$(1+x{)}^m(1-x{)}^n$

$=(1+mx+\frac{m(m-1)}{2}{x}^2+...+m{x}^{m-1}+x^m)(1-nx+\frac{n(n-1)}{2}{x}^2...+(-1{)}^{n-1}n{x}^{n-1}+(-1{)}^n{x}^n)$

Hence, the coefficient of x will be.

$m-n=3$ ...(i)

The coefficient of $x^2$ will be

$\frac{m(m-1)}{2}+\frac{n(n-1)}{2}-mn=-6$ ...(ii)

Substituting the value of m from equation (i) to equation (ii), we get.

$\frac{(n+3)(n+2)}{2}+\frac{n(n-1)}{2}-3n-n^2=-6$

$n^2+5n+6+n^2-n-6n-2n^2=-12$

$-2n+6=-12$

$-2n=-18$

$n=9$

Substituting in the equation (i) we get.

Therefore, $m=12$.