已知数列{an}的前n项和为Sn=（an+1)＾2／4,求{an}的通项an
3个回答
a1=S1=（a1+1)^2/4
4a1=(a1)^2+2a1+1
(a1)^2-2a1+1=0
(a1-1)^2=0
a1=1
Sn=（an+1)^2/4
S(n-1)=[a(n-1)+1]^2/4
an=Sn-S(n-1)
=（an+1)^2/4-[a(n-1)+1]^2/4
4an=（an+1)^2-[a(n-1)+1]^2
4an=（an)^2+2an+1-[(a(n-1)]^2-2a(n-1)-1
（an)^2-2an-[(a(n-1)]^2-2a(n-1)=0
（an)^2-[(a(n-1)]^2-2an-2a(n-1)=0
[an-a(n-1)][an+a(n-1)]-2[an+a(n-1)]=0
[an-a(n-1)-2][an+a(n-1)]=0
[an-a(n-1)-2]=0或[an+a(n-1)]=0
an-a(n-1)=2或an=-a(n-1)
an-a(n-1)=2或an/a(n-1)=-1
当a1=1,d=an-a(n-1)=2时,该数列为等差数列
an=a1+(n-1)d
=1+2(n-1)
=2n-1
当a1=1,q=an/a(n-1)=-1时,该数列为等比数列
an=a1q^(n-1)
=1*(-1)^(n-1)
=(-1)^(n-1)
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