JEE Main & Advanced Mathematics Binomial Theorem and Mathematical Induction Question Bank Self Evaluation Test - Binomial Theorem

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

A) (20, 45)

B) (35, 20)

C) (45, 35)

D) (35, 45)

Correct Answer: D

Solution :

[d] \[{{(1-y)}^{m}}{{(1+y)}^{n}}\]

\[=[1-{{\,}^{m}}{{C}_{1}}y+{{\,}^{m}}{{C}_{2}}{{y}^{2}}-....][1+{{\,}^{n}}{{C}_{1}}y+{{\,}^{n}}{{C}_{2}}{{y}^{2}}+....]\]

\[=1+(n-m)y+\left\{ \frac{m(m-1)}{2}+\frac{n(n-1)}{2}-mn \right\}{{y}^{2}}+....\]

\[=1+(n-m)y+\left\{ \frac{m(m-1)}{2}+\frac{n(n-1)}{2}-mn \right\}{{y}^{2}}+....\]By comparing coefficients with the given expression, we get

\[\therefore {{a}_{1}}=n-m=10\] and

\[{{a}_{2}}=\frac{{{m}^{2}}+{{n}^{2}}-m-n-2mn}{2}=10\]

So, \[n-m=10\] and \[{{(m-n)}^{2}}-(m+n)=20\]

\[\Rightarrow \,\,m+n=80\therefore \,\,m=35,n=45\]