设Sn是数列{an}的前n项和,a1=a,且Sn^2=3n^2an+S(n-1)^2,证明数列{a(n+2)-an}是常数数列设Sn是数列{an}的前n项和,a1=a,且Sn^2=3n^2an+S(n-1)^2,an≠0,n=2,3,4……证明数列{a(n+2)-an}(n≥2)是常数数列

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设Sn是数列{an}的前n项和,a1=a,且Sn^2=3n^2an+S(n-1)^2,证明数列{a(n+2)-an}是常数数列设Sn是数列{an}的前n项和,a1=a,且Sn^2=3n^2an+S(n-1)^2,an≠0,n=2,3,4……证明数列{a(n+2)-an}(n≥2)是常数数列

设Sn是数列{an}的前n项和,a1=a,且Sn^2=3n^2an+S(n-1)^2,证明数列{a(n+2)-an}是常数数列设Sn是数列{an}的前n项和,a1=a,且Sn^2=3n^2an+S(n-1)^2,an≠0,n=2,3,4……证明数列{a(n+2)-an}(n≥2)是常数数列
设Sn是数列{an}的前n项和,a1=a,且Sn^2=3n^2an+S(n-1)^2,证明数列{a(n+2)-an}是常数数列
设Sn是数列{an}的前n项和,a1=a,且Sn^2=3n^2an+S(n-1)^2,an≠0,n=2,3,4……证明数列{a(n+2)-an}(n≥2)是常数数列

设Sn是数列{an}的前n项和,a1=a,且Sn^2=3n^2an+S(n-1)^2,证明数列{a(n+2)-an}是常数数列设Sn是数列{an}的前n项和,a1=a,且Sn^2=3n^2an+S(n-1)^2,an≠0,n=2,3,4……证明数列{a(n+2)-an}(n≥2)是常数数列
Sn^2=3n^2an+S(n-1)^2
Sn^2-S(n-1)^2=3n^2an
[Sn+S(n-1)][Sn-S(n-1)]=3n^2an
[Sn+S(n-1)]*an=3n^2an
an≠0
所以Sn+S(n-1)=3n^2
所以S(n+1)+Sn=3(n+1)^2
相减
S(n+1)-S(n-1)=3(n+1)^2-3n^2=6n+3
同理
S(n+2)-Sn=6(n+1)+3=6n+9
相减
[S(n+2)-S(n+1)]-[Sn-S(n-1)]=6n+9-6n-3
a(n+2)-an=6
所以{a(n+2)-an}(n≥2)是常数数列

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