Binomial-Theorem-And-Its-Simple-Applications Question 218
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)!} $
Correct Answer: DShow Answer
Answer:
Solution:
$ \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
