Question

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

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

Answer 1

Answer: (C)

$\because {\left(1-y\right)}^m{\left(1+y\right)}^n=\left({}^m{C}_0{-}^m{C}_{1}y{+}^m{C}_2{y}^2-...\right)\left({}^n{C}_0{-}^n{C}_{1}y{+}^n{C}_2{y}^2+...\right)$
$\therefore a_1=$ coefficient of $y$ in ${\left(1-y\right)}^m{\left(1+y\right)}^n$
${=}^n{C}_1{-}^m{C}_1=10$
$\Rightarrow n-m=10$
$\Rightarrow n=m+10$
and $a_2=$ coefficient of $y^2$ in ${\left(1-y\right)}^m{\left(1+y\right)}^n$
${=}^n{C}_2{-}^m{C}_1{\cdot }^n{C}_1{+}^m{C}_2$
${\therefore }^n{C}_2{-}^m{C}_1{\cdot }^n{C}_1{+}^m{C}_2=10$
$\Rightarrow \frac{n\left(n-1\right)}{2}+\frac{m\left(m-1\right)}{2}-mn=10$
$\Rightarrow \frac{\left(10+m\right)\left(9+m\right)}{2}+\frac{m\left(m-1\right)}{2}$
$-m\left(10+m\right)=10\left(usingEq.\left(i\right)\right)$
$\Rightarrow 45-5m+\frac{9m}{2}-\frac{m^2}{2}+\frac{m^2}{2}-\frac{m}{2}=10$
$\Rightarrow 45-m=10\Rightarrow m=35$
$\therefore n=45\left(usingEq.\left(i\right)\right)$
$\therefore \left(m,n\right)=\left(3,45\right)$