Bernoulli trial summation by hand

[eprparadox]

Homework Statement

Show that the expected number of successes in n Bernoulli trials w probability p of success is <x> = np

Homework Equations

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)?

 

[LCKurtz]

Look here:
http://www.math.ubc.ca/~feldman/m302/binomial.pdf

 

[eprparadox]

excellent, thanks!

 

[Ray Vickson]

Homework Statement

Show that the expected number of successes in n Bernoulli trials w probability p of success is <x> = np

Homework Equations

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)?

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.