题目描述：
数列{a小n}的前n项和为S小n,a=1,a小n加1=2S小n(n属于N星号) 求数列{a小n}的通项公式a小n 急答案解析
【答案】An+1 = 2SnAn+2 = 2Sn+1
An+2 -An+1= 2(Sn+1 -Sn)=2An+1
An+2=3An+1
A1=1,A2=2S1=2
即数列An是公比是3,首项为2的等差数列.n≥2
An＝2×3^(n-2)=2*3^(n-2)
a1=1
An=2*3^(n-2),n≥2 【其他答案】a(n+1)=2Sn
Sn=a(n+1)
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