已知Sn是首项为a的等比数列{an}的前n项的和,S3,S9,S6成等差数列答:Sn是首项为a的等比数列{an}的前n项的和,则Sn=a(1-q^n)/(1-q)S3,S9,S6成等差数列,则S9-S3=S6-S9 即a[(1-q^9)-(1-q^3)]/(1-q)=a[(1-q^6)-(1-q^9)]/(1-q),q^3-q^9=q^9-q^6,所以q-q^7=q^7-q^4,aq-aq^7=aq^7-aq^4,a2-a8=a8-a5 所以a2,a8,a5...
已知数列an前n项和为Sn,首项为a1,且1,an,sn成等差数列。求an通项...答:2a(n) = s(n) + 1,2a(1) = s(1) + 1 = a(1) + 1, a(1) = 1.2a(n+1) = s(n+1) + 1,2a(n+1) - 2a(n) = s(n+1) - s(n) = a(n+1),a(n+1) = 2a(n),{a(n)}是首项为a(1)=1,公比为2的等比数列。a(n) = 2^(n-1)...