sum from 0 to N of (N choose k)*(P^k)(1-P)^k
heureka has given the result for p equals any value, not necessarily between 0 and 1. The sum isn't 1 when 0<p<1
The probabilistic expression (which is probably what was meant) is:
$$\sum_{k=0}^N(\frac{N}{k})p^{N-k}(1-p)^k$$
.
that would be 1
P is of course between 0 and 1 since it represents a probability.
$$\sum \limits_{k=0}^N {N \choose k} \cdot (P^k)(1-P)^k =[1+p\cdot(1-p)]^N$$
I do not understand your answer Heureka :/
Ok Alan, thank you :)