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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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