Question

In the expansion of $(1+x{)}^m(1-x{)}^n$, the coefficients of x and $x^2$ are respectively 3 and -6. Then m equals :

• A. 6
• B. 9
• C. 12
• D. 24

Answer 1

The correct option is C

12

$(1+x{)}^m(1-x{)}^n=(1+m{C}_{1}x+m{C}_2{x}^2+..........+m{C}_m{x}^m)(1-n{C}_{1}x+n{C}_2{x}^2-......)$

Coefficeint of x = m $C_1$ - n $C_1$ = 3

$\therefore$ m - n = 3 .........(1)

Coeffiecient of $x^2$ = $m{C}_2+n{C}_2-m{C}_1.n{C}_1=-6$

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

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

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

-2n = -18 , n = 9

$\therefore$ m = n + 3 = 12