1.设Sn是数列an的前n项和,a1=a,且Sn^2=3*n^2*an+（Sn-1)^2,an不等于0,n=2,3,4.
1.设Sn是数列an的前n项和,a1=a,且Sn^2=3*n^2*an+（Sn-1)^2,an不等于0,n=2,3,4.证明:数列{an+2-an}(n大于等于2)是常数列
2.设Sn是数列an的前n项和,且满足:4Sn=an^2+4n-1,n∈N+,⑴证明:(an -2)^2-(an-1)^2=0 ⑵满足条件的数列不唯一,试至少求出3个不同的an的通项公式
2.设Sn是数列an的前n项和,且满足:4Sn=an^2+4n-1,n∈N+,⑴证明:(an -2)^2-(an-1)^2=0 ⑵满足条件的数列不唯一,试至少求出3个不同的an的通项公式
赞 dvueiyav 幼苗
共回答了17个问题采纳率：94.1% 举报
1.Sn^2=3*n^2*an+（Sn-1)^2
[Sn+Sn-1][Sn-Sn-1]=3n^2*an
[Sn+Sn-an]*an=3n^2*an
Sn+Sn -an=3n^2
Sn=3n^2/2+an/2
S(n+1)=3(n+1)^2/2+a(n+1)/2
a(n+1)=3(2n+1)/2+a(n+1)/2-an/2
a(n+1)=6n+3-an
a(n+2)=6n+9-a(n+1)
=6n+9-6n-3+an=6+an
所以a(n+2)-an=6
2.4Sn=an^2+4n-1
4S(n-1)=a(n-1)^2+4n-5
4an=an^2-a(n-1)^2+4
an^2-4an+4-a(n-1)^2=0
(an - 2)^2-a(n-1)^2=0
4a1=a1^2+3
a1=1或3
an -2 =a(n-1)或an-2=-a(n-1)
所以an=2n-1或2n+1或1
[Sn+Sn-1][Sn-Sn-1]=3n^2*an
[Sn+Sn-an]*an=3n^2*an
Sn+Sn -an=3n^2
Sn=3n^2/2+an/2
S(n+1)=3(n+1)^2/2+a(n+1)/2
a(n+1)=3(2n+1)/2+a(n+1)/2-an/2
a(n+1)=6n+3-an
a(n+2)=6n+9-a(n+1)
=6n+9-6n-3+an=6+an
所以a(n+2)-an=6
2.4Sn=an^2+4n-1
4S(n-1)=a(n-1)^2+4n-5
4an=an^2-a(n-1)^2+4
an^2-4an+4-a(n-1)^2=0
(an - 2)^2-a(n-1)^2=0
4a1=a1^2+3
a1=1或3
an -2 =a(n-1)或an-2=-a(n-1)
所以an=2n-1或2n+1或1
1年前
5
可能相似的问题
你能帮帮他们吗