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

JEE Main · 2006 · 30 Apr 2006 · Offline · Binomial Theorem

For natural numbers

m, n

(1 - y)^{m} (1 + y)^{n} = 1 + a_{1} y + a_{2} y^{2} + \dots

, and

a_{1} = a_{2} = 10

, then

(m, n)

Answer: (D)

Expand both factors to second order:

(1 - y)^{m} = 1 - m y + (2 m) y^{2} - \dots

and

(1 + y)^{n} = 1 + n y + (2 n) y^{2} + \dots

. Multiplying, the coefficient of

y

a_{1} = n - m

and the coefficient of

y^{2}

a_{2} = (2 m) + (2 n) - mn

. Setting

a_{1} = 10

gives

n - m = 10

. Setting

a_{2} = 10

and simplifying,

\frac{m ( m - 1 )}{2} + \frac{n ( n - 1 )}{2} - mn = 10

, which rearranges to

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

. Since

(m - n)^{2} = 100

, this gives

m + n = 80

. Solving

n - m = 10

together with

m + n = 80

yields

m = 35

and

n = 45

, so

(m, n) = (35, 45)

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option