Question

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

• A. $64$
• B. $80$
• C. $88$
• D. $100$

Answer 1

The correct option is B $80$

$(1-y{)}^m(1+y{)}^n=1+a_{1}y+a_2{y}^2+...+a_{m+n}{y}^{m+n}$

Given $(a_1=a_2=10)$
$(1-my+{}^m{C}_2{y}^2+...)(1+ny+{}^n{C}_2{y}^2+...)$
$=1+a_{1}y+a_2{y}^2+....$
$\Rightarrow n-m=10\cdots \left(i\right)$
$\Rightarrow {}^m{C}_2+{}^n{C}_2-mn=10\cdots \left(ii\right)$

$\frac{m(m-1)}{2}+\frac{n(n-1)}{2}-mn=10$
$\Rightarrow \frac{m^2-m}{2}+\frac{(10+m)(9+m)}{2}-m(10+m)=10$
$\Rightarrow m^2-m+m^2+19m+90-2(m^2+10m)=20$
$\Rightarrow 18m+90-20m=20$
$\Rightarrow 2m=7$
$\therefore m=35\&n=45$
Hence, $m+n=80$