prove without bijection to the expected value of the bernoulli trials with n trials looking for an algebraic or inductive prove.
either this from abstract mathematic or discrete math, i once took these classes but now i forgot everything
Not sure if you got your answer but I'll give it a try. Haven't done any discrete math for a while... \sum_{k=0}^{n} k\binom{n}{k}p^k (1-p)^{n-k} = 0 + \sum_{k=1}^n k\binom{n}{k}p^k (1-p)^{n-k} due to the restriction on n. (We'll do this now so we don't have to worry later). \sum_{k=1}^n k\binom{n}{k}p^k (1-p)^{n-k} = \sum_{k=1}^n k\left(\frac{n!}{k!(n-k)!}\right)p^k (1-p)^{n-k} by definition of binomial coefficient. \sum_{k=1}^n k\left(\frac{n!}{k!(n-k)!}\right)p^k (1-p)^{n-k} = n! \sum_{k=1}^n \frac{p^k (1-p)^{n-k}}{(k-1)! (n-k)!} Now you see why we needed to change the lower index around. It should be clear what the next step is. You use induction to prove \sum_{k=1}^n \frac{p^k (1-p)^{n-k}}{(k-1)! (n-k)!} is equal to \frac{p}{(n-1)!}. Then you multiply this with the n! you have left over to get np. I'll write this all up in a pdf later.
So I made a mistake when I did this on paper and wound up with my approach not working. A simple search on Wikipedia yields your answer though: http://en.wikipedia.org/wiki/Binomial_distribution#Algebraic_derivations_of_mean_and_variance
