P(X=k)=0.2*5^k*e^(-5)/k!+0.8*e^(-1)/k!,k=0,1,2,3……………EX=∑P(X=k)*k=∑[0.2*5^k*e^(-5)/k!+0.8*e^(-1)/k!]*k=0.2*5+0.8*1=1.8,k从0到无穷大EX^2=∑P(X=k)*k^2=∑[0.2*5^k*e^(-5)/k!+0.8*e^(-1)/k!]*k^2=0.2*...求和是怎样求出来的？我不太懂第三步。。方法一：根据已有的结论X服从λ泊松分布的话，则有EX=DX=λ，EX^2=DX+(EX)^2=EX+(EX)^2=λ+λ^2方法二：用微积分的方法求解