在概率论中,知道x的概率满足几何分布,即P(x=k)=p*[q^(k-1)],求E(X^2)怎么求的啊

E(X^2)
=p+2^2*qp+3^2*q^2*p+……+k^2*q^(k-1)*p+……
=p(1+2^2*q+3^2*q^2+……+k^2*q^(k-1)+……)
对于上式括号中的式子,利用导数,关于q求导：k^2*q^(k-1)=(k*q^k)',并用倍差法求和,有
1+2^2*q+3^2*q^2+……+k^2*q^(k-1)+……
=(q+2*q^2+3*q^3+……+k*q^k+……)'
=[q/(1-q)^2]'
=[(1-q^2)+2(1-q)q]/(1-q)^4
=(1-q^2)/(1-q)^4
=(1+q)/(1-q)^3
=(2-p)/p^3
因此E(X^2)=p[(2-p)/p^3]=(2-p)/p^2

必须说明一点，K不可能取值0，所以累加应该是从K=(1->+∞)
令S(x)=∑k²(1-x)x^(k-1) |x|<1
S(x)=(1-x)∑k²x^(k-1)
=(1-x)∑[kx^k]'
=(1-x)[x∑kx^(k-1)]'
=(1-x)[x∑[x^k]']'
=(1-x)[x[∑x^k]']'
=(1...