已知：an=3n-1，bn=2^n，求数列{anbn}的前n项和

cn=anbn=(3n-1)*2^n
Sn=2*2^1+5*2^2+……+(3n-1)*2^n
2Sn= 2*2^2+……+(3n-4)*2^n+(3n-1)*2^(n+1)
相减：
Sn=(3n-1)*2^(n+1)-3*(2^2+2^3+……+2^n)-2*2^1
=(3n-1)*2^(n+1)-3*[2^2-2^(n+1)]/(1-2)-4
=(3n-1)*2^(n+1)-3*2^(n+1)+12-4
=(3n-4)*2^(n+1)+8

anbn=(3n-1)*2^n
Sn=(3*1-1)*2^1+(3*2-1)*2^2+(3*3-1)*2^3+……+[3*(n-1)-1]*2^(n-1)+(3*n-1)*2^n
则2Sn=(3*1-1)*2^1*2+(3*2-1)*2^2*2+(3*3-1)*2^3*2+……+[3*(n-1)-1]*2^(n-1)*2+(3*n-1)*2^n*2
=(3*1-1)*2^2+(3*2-1)*2^3+(3*3-1)*2^4+……+[3*(n-1)-1]*2^n+(3*n-1)*2^(n+1)
下式减去上式：
Sn=-(3*1-1)*2^1-3*2^2-3*2^3-3*2^4-……-3*2^n+(3*n-1)*2^(n+1)
=-4+(3*n-1)*2^(n+1)-3*(2^2+2^3+2^4+……+2^n)
=-4+(3*n-1)*2^(n+1)-3*4*(2^n-1)
=8+3*n*2^(n+1)-7*2^(n+1)