• # question_answer For  natural   numbers   m,   n,  if ${{(1-y)}^{m}}{{(1+y)}^{n}}=1+{{a}_{1}}y+{{a}_{2}}{{y}^{2}}+.....$and${{a}_{1}}={{a}_{2}}=10,$then (m, n) is     AIEEE  Solved  Paper-2006 A) (35, 20)          B)        (45, 35) C)        (35, 45)             D)        (20, 45)

Given, ${{(1-y)}^{m}}{{(1+y)}^{n}}=1+{{a}_{1}}y+{{a}_{2}}{{y}^{2}}+.....$ Since,${{(1-y)}^{m}}{{(1+y)}^{n}}={{(}^{m}}{{C}_{0}}{{-}^{m}}{{C}_{1}}y{{+}^{m}}{{C}_{2}}{{y}^{2}}-....)$ $\times {{(}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}y{{+}^{n}}{{C}_{2}}{{y}^{2}}+....)$  $\therefore$${{a}_{1}}=$Coefficient of y in ${{(1-y)}^{m}}{{(1+y)}^{n}}$ ${{=}^{n}}{{C}_{1}}{{-}^{m}}{{C}_{1}}=10$ $\Rightarrow$$n-m=10$ $\Rightarrow$$n=m+10$                                  ...(i) and${{a}_{2}}=$Coefficient of${{y}^{2}}$in ${{(1-y)}^{m}}{{(1+y)}^{n}}$ ${{=}^{n}}{{C}_{2}}{{-}^{m}}{{C}_{1}}{{.}^{n}}{{C}_{1}}{{+}^{m}}{{C}_{2}}$ $\therefore$$^{n}{{C}_{2}}{{-}^{m}}{{C}_{1}}{{.}^{n}}{{C}_{1}}{{+}^{m}}{{C}_{2}}=10$ $\Rightarrow$$\frac{n(n-1)}{2}+\frac{m(m-1)}{2}-mn=10$ $\Rightarrow$$\frac{(10-m)(9+m)}{2}+\frac{m(m-1)}{2}-m(10+m)=10$ [from Eq. (i)] $\Rightarrow$$45+\frac{19m}{2}+\frac{{{m}^{2}}}{2}+\frac{{{m}^{2}}}{2}-\frac{m}{2}-10m-{{m}^{2}}=10$ $\Rightarrow$$45-m=10\Rightarrow m=35$ $\therefore$  $n=45$                                               [from Eq. (i)] Hence,                 $(m,\text{ }n)=(35,\text{ }45)$ Alternate Solution ${{(1-y)}^{m}}{{(1+y)}^{n}}=1+{{a}_{1}}y+{{a}_{2}}{{y}^{2}}+a{{ & }_{3}}{{y}^{3}}+.....$ On differentiating w.r.t. y. we get $-m{{(1-y)}^{m-1}}{{(1+y)}^{n}}+{{(1-y)}^{m}}n{{(1+y)}^{n-1}}$ $={{a}_{1}}+2{{a}_{2}}y+3{{a}_{3}}{{y}^{3}}+...$                                    ... (i) Now, putting$y=0$in Eq. (i), we get $-m+n={{a}_{1}}=10$                             ...(ii) where, ${{a}_{1}}=10$ Again, differentiating Eq. (i), we get $-m+[(m-1){{(1-y)}^{m-2}}{{(1+y)}^{2}}{{(1-y)}^{m-1}}n{{(1+y)}^{n-1}}]$ $+n[-m{{(1-y)}^{m-1}}{{(1+y)}^{n-1}}+{{(1-y)}^{m}}(n-1){{(1+y)}^{n-2}}]$ $=2{{a}_{2}}+6{{a}_{3}}+y+...$                                ...(iii) Now, putting y = 0 in Eq. (iii), we get $-m[-(m-1)+n]+n[-m+(n-1)]=2{{a}_{2}}=20$ $\Rightarrow$$m(m-1)-mn-mn+n(n-1)=20$ $\Rightarrow$${{m}^{2}}+{{n}^{2}}-m-n-2mn=20$ $\Rightarrow$${{(m-n)}^{2}}-(m+n)=20$ $\Rightarrow$$100-(m+n)=20$                                                 $(\because -m+n=10)$ $\Rightarrow$               $m+n=80$                            ...(iv) On solving Eqs. (ii) and (iv), we get $m=35$and $n=45$