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:
N∑k=0(Nk)pN−k(1−p)k
.
that would be 1
P is of course between 0 and 1 since it represents a probability.
N∑k=0(Nk)⋅(Pk)(1−P)k=[1+p⋅(1−p)]N
I do not understand your answer Heureka :/
Ok Alan, thank you :)