用英语写证明题

Answer. The probability that X=k is C_n^k p^k q^{n-k}, so we getEX= \sum_0^n k C_n^k p^k q^{n-k} = p(\sum_0^n C_n^k p^k q^{n-k})_p= p ((p+q)^n)_p= p n(p+q)^{n-1}= npHere _p means taking partial derivative with respect to the variable p.Similarly, you can work out \sum_0^n k(k+1) C_n^k p^k q^{n-k}, and deduce the value of \sum_0^n k^2 C_n^k p^k q^{n-k}, then Var X can be computed out. But I compute out Var X = n^2 pq, rather npq. This should be checked.