已知数列an满足a1=1 Sn=2an+n 求an

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已知数列an满足a1=1 Sn=2an+n 求an
已知数列an满足a1=1 Sn=2an+n 求an

已知数列an满足a1=1 Sn=2an+n 求an
n=1时,a1=S1=2a1+1
a1=-1
n≥2时,an=Sn-S(n-1)=2an+n-[2a(n-1)+(n-1)]
an=2a(n-1)-1
an-1=2a(n-1)-2=2[a(n-1)-1]
(an -1)/[a(n-1)-1]=2,为定值
a1-1=-1-1=-2,数列{an-1}是以-2为首项,2为公比的等比数列
an -1=(-2)·2^(n-1)=-2ⁿ
an=1-2ⁿ
数列{an}的通项公式为an=1-2ⁿ

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