a2=4,a1+a2=6
a1=2,q=a2/a1=2
an=2^n
设Cn=an * bn
a1b1 ＋a2b2＋……＋anbn＝2+(n-1)2^(n＋1)
a1b1 ＋a2b2＋……＋an-1bn-1= 2+(n-2)2^n
两式相减得：anbn=[2(n-1)-(n-2)]*2^n
= n *2^n
=n*an
所以,bn=n
数列bn的前n项和:Sn=n(n+1)/2