If P_n denotes the product of the binomial coefficients in the expansion of (1+x)^n , then (P_(n+1) /P


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Question Number 52600 by Tip Top last updated on 10/Jan/19

$If P_{n} denotes the product of the binomial$ $coefficients in the expansion of {(1 + x)}^{n},$ $then \frac{P_{n + 1}}{P_{n}} equals$
Commented by Abdo msup. last updated on 10/Jan/19

$w e h a v e {(x + 1)}^{n} = \sum_{k = 0}^{n} C_{n}^{k} x^{k} \Rightarrow$ $P_{n} = \prod_{k = 0}^{n} C_{n}^{k} = C_{n}^{o} . C_{n}^{1} . . . . . C_{n}^{n} \Rightarrow$ $P_{n + 1} = C_{n + 1}^{0} . C_{n + 1}^{1} . . . C_{n + 1}^{n + 1} \Rightarrow$ $\frac{P_{n + 1}}{P_{n}} = \frac{\prod_{k = 0}^{n} C_{n}^{k}}{\prod_{k = 0}^{n + 1} C_{n + 1}^{k}} = \frac{\prod_{k = 0}^{n} C_{n}^{k}}{\prod_{k = 0}^{n} \frac{(n + 1)!}{k! (n + 1 - k)!}}$ $= \frac{\prod_{k = 0}^{n} C_{n}^{k}}{\prod_{k = 0}^{n} \frac{n + 1}{(n + 1 - k)} \prod_{k = 0}^{n} \frac{n!}{k! (n - k)!}}$ $= \prod_{k = 0}^{n} \frac{(n + 1 - k)}{n + 1} = \frac{(n + 1)!}{{(n + 1)}^{n + 1}} = \frac{(n + 1) n!}{(n + 1) {(n + 1)}^{n}}$ $= \frac{n!}{{(n + 1)}^{n}} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

	$p_{n} = {n c}_{0} \times {n c}_{1} \times {n c}_{2} . . . {n c}_{n} \leftarrow p r o d u c t o f n t e r m s$ $p_{n + 1} = n + 1 c_{0} \times n + 1 c_{1} \times n + 1 c_{2} . . \times n + 1 c_{n + 1} \leftarrow p r o d u c t o f (n + 1) t e r m s$ $n o w r a t i o o f r + 1 t h t e r m$ $= \frac{n + 1 c_{r}}{{n c}_{r}}$ $= \frac{\frac{(n + 1)!}{r! (n + 1 - r)!}}{\frac{n!}{r! (n - r)!}} = \frac{(n + 1)!}{n!} \times \frac{(n - r)!}{(n + 1 - r)!}$ $= \frac{n + 1}{n + 1 - r}$ $s o \frac{p_{n + 1}}{p_{n}} . = \frac{}{}$ $= \frac{n + 1}{n + 1 - 0} \times \frac{n + 1}{n + 1 - 1} \times \frac{n + 1}{n + 1 - 2} . . .$ $= \frac{{(n + 1)}^{n}}{(n + 1)!}$
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