show that Σ_(x=0) ^n xp(x)=np given that p(x)=^n C


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Question Number 50856 by peter frank last updated on 21/Dec/18

$s h o w t h a t$ $\underset{x = 0}{\sum^{n}} x p (x) = n p g i v e n$ $t h a t p (x) =^{n} C_{x} p^{x} q^{n - x}$
	Answered by Smail last updated on 21/Dec/18

	$Q (p) = \underset{x = 0}{\sum^{n}} p (x) = \underset{x = 0}{\sum^{n}}^{n} C_{x} p^{x} q^{n - x} = {(p + q)}^{n}$ $\frac{d Q (p)}{d p} = \underset{x = 0}{\sum^{n}}^{n} C_{x} ({x p}^{x - 1}) q^{n - x} = n {(p + q)}^{n - 1}$ $= \underset{x = 0}{\sum^{n}} x^{n} C_{x} p^{x} q^{n - x} \times \frac{1}{p} = n {(p + q)}^{n - 1}$ $= \underset{x = 0}{\sum^{n}} x (^{n} C_{x} p^{x} q^{n - x}) = n {(p + q)}^{n - 1} p$ $\underset{x = 0}{\sum^{n}} x p (x) = n p {(p + q)}^{n - 1}$
	Commented by Smail last updated on 21/Dec/18

	$I t h i n k y o u a r e m i s s i n g s o m e t h i n g .$
	Commented by peter frank last updated on 21/Dec/18

	$t h a n k y o u$
	Commented by Smail last updated on 22/Dec/18

	$Y o u a r e w e l c o m e$
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