Question
For natural numbers $$m, n$$ if $${\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n} = 1 + {a_1}y + {a_2}{y^2} + .....$$ and $${a_1} = {a_2} = 10,$$ then $$\left( {m,n} \right)$$ is
A.
$$\left( {20,45} \right)$$
B.
$$\left( {35,20} \right)$$
C.
$$\left( {45,35} \right)$$
D.
$$\left( {35,45} \right)$$
Answer :
$$\left( {35,45} \right)$$
Solution :
$$\eqalign{
& {\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n} \cr
& = \left[ {1 - {\,^m}{C_1}y + {\,^y}{C_2}{y^2} - .....} \right]\left[ {1 + {\,^n}{C_1}y + {\,^n}{C_2}{y^2} + .....} \right] \cr
& = 1 + \left( {n - m} \right)y + \left\{ {\frac{{m\left( {m - 1} \right)}}{2} + \frac{{n\left( {n - 1} \right)}}{2} - mn} \right\}{y^2} + ..... \cr} $$
By comparing coefficients with the given expression, we get
$$\eqalign{
& \therefore {a_1} = n - m = 10{\text{ and }}{a_2} = \frac{{{m^2} + {n^2} - m - n - 2mn}}{2} = 10 \cr
& {\text{So, }}n - m = 10{\text{ and }}{\left( {m - n} \right)^2} - \left( {m + n} \right) = 20 \cr
& \Rightarrow m + n = 80 \cr
& \therefore m = 35,n = 45 \cr} $$