Binomial Theorem And Its Simple Applications Question 3
Question: For natural numbers $ m,nif{{(1-y)}^{m}}{{(1+y)}^{n}} $ $ =1+a_1y+a_2y^{2}+… $ and $ a_1=a_2=10 $ , then $ (m,n) $ is
Options:
A) (20, 45)
B) (35, 20)
C) (45, 35)
D) (35, 45)
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] $ {{(1-y)}^{m}}{{(1+y)}^{n}} $
$ =[1-{{,}^{m}}C_1y+{{,}^{m}}C_2y^{2}-….][1+{{,}^{n}}C_1y+{{,}^{n}}C_2y^{2}+….] $
$ =1+(n-m)y+{ \frac{m(m-1)}{2}+\frac{n(n-1)}{2}-mn }y^{2}+…. $
$ =1+(n-m)y+{ \frac{m(m-1)}{2}+\frac{n(n-1)}{2}-mn }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 $
BETA
