已知数列{an}的前n项和Sn=n2,数列{bn}是等比数列,且a1=b1,b2（a2-a1）=b1

sn=n²,s1=1,a1=1
an=sn-s(n-1)=n²-(n-1)²=2n-1
b1=1,b2=b1/(a2-a1)=1/2
bn=(1/2)^(n-1)
cn=(2n-1)/2^(n-1)=n/2^(n-2)-1/2^(n-1)
tn=1/2^-1+2/2^0+3/2^1+.+n/2^(n-2)-(1/2^0+...+1/2^(n-1))
tn=1/2^-1+2/2^0+3/2^1+.+n/2^(n-2)-(2-1/2^(n-1))
tn/2=1/2^0+2/2^1+...+n/2^(n-1)-(1/2^1+..+1/2^n)
tn/2=1/2^0+2/2^1+...+n/2^(n-1)-(1-1/2^n)
tn-tn/2=1/2^-1+1/2^0+1/2^1+...+1/2^(n-2)-n/2^(n-1)-1-1/2^n
tn/2=2+1+1-1/2^(n-2)-n/2^(n-1)-1-1/2^n
tn/2=3-5/2^(n-2)-n/2^(n-1)
tn=6+(5-2n)/2^(n-1)
tn+cn=6-4/2^(n-1)
tn+ca=6-1/2^(n-3)