Binomial Theorem And Its Simple Applications Question 27
Question: If $ \pi (n) $ denotes product of all binomial coefficients in $ {{(1+x)}^{n}} $ then ratio of $ \pi (2002) $ to $ \pi (2001) $ is
Options:
A) 2002
B) $ \frac{{{(2002)}^{2001}}}{(2001)!} $
C) $ \frac{{{(2001)}^{2002}}}{(2002)!} $
D) 2001
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] $ \frac{\pi (n)}{\pi (n+1)}=\frac{^{n}C_0{{.}^{n}}C_1{{.}^{n}}C_2{{…}^{n}}C_{n}}{^{n+1}C_0{{.}^{n+1}}C_1{{.}^{n+1}}C_2{{…}^{n+1}}{C_{n+1}}} $
$ =\frac{1}{^{n+1}C_0}( \frac{^{n}C_0}{^{n+1}C_1} )( \frac{^{n}C_1}{^{n+1}C_2} )…..( \frac{^{n}C_{n}}{^{n+1}{C_{n+1}}} ) $
$ =\frac{1}{1}( \frac{1}{n+1} )( \frac{2}{n+1} )……( \frac{n+1}{n+1} ) $
$ [ \because ,\frac{^{n}C_{r}}{^{n+1}{C_{r+1}}}=\frac{r+1}{n+1} ] $
$ =\frac{(n+1)!}{{{(n+1)}^{n+1}}}=\frac{n!}{{{(n+1)}^{n}}} $
$ \therefore ,\frac{\pi (2002)}{\pi (2001)}=\frac{{{(2002)}^{2001}}}{(2001)!} $
BETA
