an+2-an+1=1/2(an+1-an)
dn+1=dn/2
dn+1/dn=1/2
{dn}是公比为1/2的等比数列
d1=1/2
{dn}前n项合计为Sn
Sn=1-1/2^n
Sn=a2-a1+a3-a2+a4-a3……an+1-an=an+1-a1=an+1-1
1-1/2^n=an+1-1
an+1=2-1/2^n
an=2-1/2^(n-1)
令cn=an*bn
cn=(3n-2)-(3n-2)/2^(n-1)
设Sn=S1-S2
S1=3(1+2+3+……n)-2n=(3n^2-n)/2
S2=1/2^0+4/2^1+7/2^2……(3n-2)/2^(n-1)
S2/2=1/2^1+4/2^2+7/2^3……(3n-2)/2^n
前式减后式
S2/2=1/2^0+3/2^1+3/2^2……3/2^(n-1)-(3n-2)/2^n
=1-(3n-2)/2^n+3[1/2+1/2^2+……1/2^(n-1)]
=4-3/2^(n-1)-(3n-2)/2^n
S2=8-3/2^(n-2)-(3n-2)/2^(n-1)
Sn=S1-S2=(3n^2-n)/2-8+3/2^(n-2)+(3n-2)/2^(n-1)