Question

已知数列{an}的前n项和为Sn=（an+1)＾2／4,求{an}的通项an

Answer 1

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)

Answer 2

http://zhidao.baidu.com/question/135095262.html?si=1
先百度 ，再提问

Answer 3

an=Sn-S(n-1)
=1/4(an+1)2-1/4[a(n-1)+1]2
an=1/4(an2+2an+1)-1/4[a(n-1)+1]2
1/4(an2+2an+1)-an=1/4[a(n-1)+1]2
1/4(an2-2an+1)=1/4[a(n-1)+1]2
1/4(an-1)2=1/4[a(n-1)+1]2
所以an-1=a(n-1)+1
或an-1=-[a(n-1)+1]

若an-1=-[a(n-1)+1]
an=-a(n-1)
an是正数，所以不成立

所以an-1=a(n-1)+1
an-a(n-1)=2
所以是等差数列
d=2
a1=S1
所以a1=1/4(a1+1)2
(a1-1)2=0
a1=1

所以an=2n-1