Question

For natural numbers m, n if $(1-y{)}^m(1+y{)}^n$
$=21+a_{1}y+a_2{y}^2+...$ and $a_1=a_2=10$, then (m, n) is

• A. (20, 45)
• B. (35, 20)
• C. (45, 35)
• D. (35, 45)

Answer 1

Answer: (D)

${\left(1-y\right)}^m{\left(1+y\right)}^n$
$=\left[1{-}^m{C}_{1}y{+}^m{C}_2{y}^2-.....\right]\left[1{+}^n{C}_{1}y{+}^n{C}_2{y}^2+.....\right]$
$=1+\left(n-m\right)y+\left\{\frac{m\left(m-1\right)}{2}+\frac{n\left(n-1\right)}{2}-mn\right\}{y}^2+.....$
By comparing coefficients with the given expression, we get
$\therefore a_1=n-m=10$ and
$a_2=\frac{m^2+n^2-m-n-2mn}{2}=10$
So, $n-m=10$ and ${\left(m-n\right)}^2-\left(m+n\right)=20$
$\Rightarrow m+n=80\therefore m=35,n=45$