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 :)