1. Tardigrade
  2. Question
  3. Mathematics
  4. If Pn denotes the product of the binomial coefficients in the expansion of (1+x)n, then (Pn+1/Pn) equals

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If $P_n$ denotes the product of the binomial coefficients in the expansion of $\left(1+x\right)^{n}$, then $ \frac{P_{n+1}}{P_{n}} $ equals

Binomial Theorem

Solution:

Here, $P_{n} = ^{n}C_{0}. ^{n}C_{1}. ^{n}C_{2} .... ^{n}C_{n}$
and $P_{n+1} = ^{n+1}C_{0}. ^{n+1}C_{1}. ^{n+1}C_{2} .... ^{n+1}C_{n+1}$
$\therefore \frac{P_{n+1}}{P_{n}} = \frac{^{n+1}C_{0}. ^{n+1}C_{1}. ^{n+1}C_{2} .... ^{n+1}C_{n+1}}{^{n}C_{0}. ^{n}C_{1}. ^{n}C_{2} .... ^{n}C_{n}}$
$= \left( \frac{^{n+1}C_{1}}{^{n}C_{0}}\right)\left( \frac{^{n+1}C_{2}}{^{n}C_{1}}\right)\left( \frac{^{n+1}C_{3}}{^{n}C_{2}}\right) ..... \left(\frac{^{n+1}C_{n+1}}{^{n}C_{n}}\right)$
$= \left(\frac{n+1}{1}\right)\left(\frac{n+1}{2}\right)\left(\frac{n+1}{3}\right).....\left(\frac{n+1}{n+1}\right) = \frac{\left(n+1\right)^{n+1}}{\left(n+1\right)\,!}$