For natural numbers m,n, if (1-y)^m(1+y)^n=1+a_1y+a_2y^2+ and a_1=a_2=10, then (m,n) is:

Question

For natural numbers $m,n$, if $(1-y)^m(1+y)^n=1+a_1y+a_2y^2+\cdots$ and $a_1=a_2=10$, then $(m,n)$ is:

Accepted Answer

$$(1-y)^m(1+y)^n = \left[1-{}^mC_1y+{}^mC_2y^2-\cdots\right]\left[1+{}^nC_1y+{}^nC_2y^2+\cdots\right]$$

$$=1+(n-m)y+\left[\frac{m(m-1)}2+\frac{n(n-1)}2-mn\right]y^2+\cdots$$

$a_1=n-m=10$; $a_2=\dfrac{(m-n)^2-(m+n)}{2}=10\Rightarrow100-(m+n)=20\Rightarrow m+n=80$

Solving $n-m=10$, $m+n=80$: $m=35,\ n=45$

Solutions · 1