已知数列{an},前n项和Sn=3/2(an-1),求通项an . n是在下标.
人气:186 ℃ 时间:2019-11-13 02:24:50
解答
n=1时a1=s1,代入Sn=3/2(an-1)得a1=3/2(a1-1)解得a1=S1=3.
当n>=2时,an=Sn-Sn-1,代入Sn=3/2(an-1)整理得:Sn=3Sn-1 +3
即有:Sn+3/2=3(Sn-1 +3/2),故数列{Sn+3/2}是一个等比数列
Sn+3/2=(S1+3/2)*3^n-1,由此可得:Sn=3/2 (3^n-1)
an=Sn-Sn-1=3/2 (3^n-1)-3/2(3^n-1 -1)=3^n (n>=2)
即an=3^n (n>=2) 而n=1时a1=3 也满足an=3^n
故an的通项公式为:an=3^n
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