If $\mathrm P(X=k)=\binom nkp^k(1-p)^{n-k}$ for a binomial distribution, then from the definition of the expected value $$\mathrm E(X) = \sum^n_{k=0}k\mathrm P(X=k)=\sum^n_{k=0}k\binom nkp^k(1-p)^{n-k}$$ but the expected value of a Binomal distribution is $np$, so how is
$$\sum^n_{k=0}k\binom nkp^k(1-p)^{n-k}=np$$