Bernoulli trial summation by hand

1. Aug 27, 2014

eprparadox

1. The problem statement, all variables and given/known data
Show that the expected number of successes in n Bernoulli trials w probability p of success is <x> = np

2. Relevant equations

3. The attempt at a solution

So I get the right answer which is this: $$E\left( x\right) =\sum _{x=0}^{n}x\left( \begin{matrix} n\\ x\end{matrix} \right) p^{x}\left( 1-p\right) ^{n-x}$$

I know it's right, though, because I inputted it into wolframalpha and get <x> = np.

My question is how do I try to sum this by hand. I really have no idea where to start and I'm curious as to how this is done. Do I maybe use Stirling's formula to simplify the C(n, x)?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 27, 2014

LCKurtz

3. Aug 27, 2014

eprparadox

excellent, thanks!

4. Aug 27, 2014

Ray Vickson

This was already answered for you in another thread: see post #5 in your thread on the expected number of 5's in tossing a die.

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