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For natural numbers $m,n$ ,if ${\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n} = 1 + {a_1}y + {a_2}{y^2} + \ldots \;$ and $a_1= a_2=10,$ then $(m,n)$ =______.
$(20,45)$
$(35,20)$
$(45,35)$
$(35,45)$
Solution
$(1-y)^{m}(1+y)^{n}$
$=\left(^{m} C_{0}-^{m} C_{1} y+^{m} C_{2} y^{2}+\ldots .\right)$$\left(^{n} C_{0}+^{n} C_{1} y+^{n} C_{2} y^{2}+\ldots\right)$
$a_{1}=$ Coefficient of $y=^{n} C_{1}-^{m} C_{1}=10$
$\Rightarrow n-m=10$
$a_{2}=$ Coefficient of $y^{2}$
$=^{n} C_{2}+^{n} C_{1} \times^{m} C_{1}+^{m} C_{2}=10$
$\Rightarrow \frac{n(n-1)}{2}-n m+\frac{m(m-1)}{2}=10$
$\Rightarrow n(n-1)-2 n m+m(m-1)=20$
$\Rightarrow(m+10)(m+9)-2(m+10) m+m(m-1)=20$
$\Rightarrow 90+19 m+m^{2}-2 m^{2}-20 m+m^{2}-m-20=0$
$\Rightarrow 70-2 m=0$
$\Rightarrow m=35$
$\therefore n=10+35=45$
