Question

If $(1-y{)}^{\prime \prime \prime }\cdot (1+y{)}^{\prime \prime }=1+a_{1}y+a_2{y}^2+...+a_{m+n}{y}^{m+n}$ where $m\in N$ 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)

$(35,45)$

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

$a_1=a_2=10$

Differentiate on both sides w.r.t y

$-m(1-y{)}^{m-1}(1+y{)}^n+n(1-y{)}^m(1+y{)}^{n-1}=a_1+2a_{2}y+......+(m+n)a_{m+n}{y}^{m+n-1}$ ....... $\left(1\right)$

Put y=0,

$\therefore n-m=a_1=10$

$\therefore n-m=10$

Differentiate (1) w.r.t $y$, we get

$m(m-1)(1-y{)}^{m-2}(1+y{)}^n-2mn(1-y{)}^{m-1}(1+y{)}^{n-1}+(n-1)\left(n\right)(1-y{)}^m(1+y{)}^{n-2}$

$=2a_2+6a_{3}y+.....+(m+n)(m+n-1)a_{m+n}{y}^{m+n-2}$

Put $y=0,$

$m(m-1)-2mn+n(n-1)=2a_2$

$(m-n{)}^2-(m+n)=20$

$\therefore m+n=80$

$n-m=10$

$m+n=80$

$\therefore n=45,m=35$

$(m,n)=(35,45)$