Binomial Theorem And Its Simple Applications Question 54
Question: If $ P_{n} $ denotes the product of the binomial coefficients in the expansion of $ {{(1+x)}^{n}} $ , then $ \frac{{P_{n+1}}}{P_{n}} $ equals
Options:
A) $ \frac{n+1}{n!} $
B) $ \frac{n^{n}}{n!} $
C) $ \frac{{{(n+1)}^{n}}}{(n+1)!} $
D) $ \frac{{{(n+1)}^{n+1}}}{(n+1)!} $
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Answer:
Correct Answer: D
Solution:
- [d] Here, $ P_{n}={{,}^{n}}C_0.{{,}^{n}}C_1.{{,}^{n}}C_2….{{,}^{n}}C_{n} $ and $ {P_{n+1}}={{,}^{n+1}}C_0.{{,}^{n+1}}C_1.{{,}^{n+1}}C_2….{{,}^{n+1}}{C_{n+1}} $
$ \therefore ,\frac{{P_{n+1}}}{P_{n}}=\frac{^{n+1}C_0.{{,}^{n+1}}C_1.{{,}^{n+2}}C_2….{{,}^{n+1}}{C_{n+1}}}{^{n}C_0.{{,}^{n}}C_1.{{,}^{n}}C_2….{{,}^{n}}C_{n}} $
$ =( \frac{^{n+1}C_1}{^{n}C_0} )( \frac{^{n+1}C_3}{^{n}C_1} )( \frac{^{n+1}C_3}{^{n}C_0} )….( \frac{^{n+1}{C_{n+1}}}{^{n}C_{n}} ) $
$ =( \frac{n+1}{1} )( \frac{n+1}{2} )( \frac{n+1}{3} )….( \frac{n+1}{n+1} ) $
$ =\frac{{{(n+1)}^{n+1}}}{(n+1)!} $
BETA
