Binomial Theorem And Its Simple Applications Question 9
Question: If $ \frac{e^{x}}{1-x}=B_0+B_1x+B_2x^{2}+…+B_{n}x^{n} $ then $ B_{n}-{B_{n-1}} $ is
Options:
A) $ \frac{1}{n!}-\frac{1}{(n-1)!} $
B) $ \frac{1}{n!} $
C) $ \frac{1}{(n-1)!} $
D) $ \frac{1}{n!}+\frac{1}{(n-1)!} $
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] We have $ (1-x)(B_0+B_1x+B_2x^{2}+….) $
$ …..+{B_{n-1}}{x^{n-1}}+B_{n}x^{n}+…) $
$ =e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+….+\frac{x^{n}}{n!}+…….(1) $
Hence equating the coefficients of $ x^{n} $ on both sides of (1) we get $ B_{n}-{B_{n-1}}=\frac{1}{n!}. $
BETA
