Question

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

• 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 $\left(1 - y\right)^{m} \left(1 + y\right)^{n}$
$=^{n}C_{2}-^{m}C_{1}\cdot^{n}C_{1}+^{m}C_{2}$
$\therefore ^{n}C_{2}-^{m}C_{1}\cdot ^{n}C_{1}+^{m}C_{2}=10$
$\Rightarrow \frac{n\left(n-1\right)}{2}+\frac{m\left(m-1\right)}{2}-mn=10$
$\Rightarrow \frac{\left(10+m\right)\left(9+m\right)}{2}+\frac{m\left(m-1\right)}{2}$
$-m\left(10+m\right)=10 \left(using Eq. \left(i\right)\right)$
$\Rightarrow 45-5m+\frac{9m}{2}-\frac{m^{2}}{2}+\frac{m^{2}}{2}-\frac{m}{2}=10$
$ \Rightarrow 45- m = 10 \Rightarrow m = 35$
$\therefore n = 45 \left(using Eq. \left(i\right)\right)$
$\therefore \left(m, n\right) = \left(3, 45\right)$