Question

If the coefficients of $x$ and $x^2$ in the expansion of ${(1+x)}^m{(1-x)}^n$ are $3$ and $-6$ respectively. Find the values of $m$ and $n$.

Answer 1

We have

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

$=[1{+}^m{C}_{1}x{+}^m{C}_2{x}^2{+}^m{C}_3{x}^3+.....][1{-}^n{C}_{1}x{+}^n{C}_2{x}^2{-}^n{C}_3{x}^3+....]$

Coefficient of x will be obtained when a constant term in the first bracket is multiplied with term having x in second bracket

Given that coefficient of $x=3$

${\Rightarrow }^m{C}_1{-}^n{C}_1=3$

$\Rightarrow m-n=3$

$\Rightarrow m=n+\mathrm{3.....}\left(1\right)$

Coefficient of $x^2=-6$

${\Rightarrow }^m{C}_2{+}^n{C}_2{-}^m{C}_1{.}^n{C}_1=-6$

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

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

$\Rightarrow (n+3)(n+2)+n(n-1)-2(n+3)n=-12$

$\Rightarrow n^2+5n+6+n^2-n-2n^2-6n+12=0$

$\Rightarrow 2n^2-2n^2-2n+18=0$

$\Rightarrow 2n=18$

$\Rightarrow n=9$

Putting in (1)

$m=n+3$

$m=9+3$

$m=12$

So the value of m and n are 12 and 9