设等差数列an=a1+(n-1)d
等比数列bn=b1q^(n-1)
其积cn=anbn,cn的和为Sn
Sn=a1b1+a2b2+...+anbn
qSn= a1b2+...+a(n-1)bn+anb(n+1)
两式相减：（1-q)Sn=a1b1+db2+...+dbn-anb(n+1)=a1b1+d(b2+...bn)-anb(n+1)=a1b2+db2[1-q^(n-1)]/(1-q)- anb(n+1)
因此Sn=a1b2/(1-q)+db2[1-q^(n-1)]/(1-q)^2-anb(n+1)/(1-q)