If S ~B(n,p), then E(S) = np.
Proof: Since S~B(n,p), S = + +
where each of the Xi ~ Ber(p) and E(Xi) = p.
Therefore,
E(S) = E(+ + )
= E(X1) + E(X2) + …. + E(Xn)
= p + p + p + …. + p = np.