Binomial Theorem And Its Simple Applications Question 90
Question: Let $ f(x)=a_0+a_1x+a_2x^{2}+…+a_{n}x^{n}+… $ and $ \frac{f(x)}{1-x}=b_0+b_1x+b_2x^{2}+…+b_{n}x^{n}+…., $ then
Options:
A) $ b_{n}={b_{n-1}}=a_{n} $
B) $ b_{n}-{b_{n-1}}=a_{n} $
C) $ b_{n}/{b_{n-1}}=a_{n} $
D) none of these
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] $ \frac{f(x)}{1-x}=b_0+b_1x+b_2x^{2}+…+b_{n}x^{n}+… $
$ \Rightarrow a_0+a_1x+a_2x^{2}+…+a_{n}x^{n}+… $
$ =(1-x)(b_0+b_1x+b_2x^{2}+…+b_{n}x^{n}+…) $ Comparing the coefficient of $ x^{n} $ on both the sides, $ a_{n}=b_{n}-{b_{n-1.}} $
BETA
