已知Sn是数列｛an｝的前n项和,求证:若Sn=n/2*(an+a1),则｛an｝是等差数列,反之也成立

Question

已知Sn是数列｛an｝的前n项和,求证:若Sn=n/2*(an+a1),则｛an｝是等差数列,反之也成立

Answer 1

先证明：若Sn=n/2*(an+a1),则{an}是等差数列
a1=S1=1/2*(a1+a1)=a1
对于n>=2,有
an=Sn-S(n-1)=n/2*(an+a1)-(n-1)/2*[a(n-1)+a1]=n/2*an-(n-1)/2*a(n-1)+1/2*a1
所以an=a1+(n-1)[an-a(n-1)]
于是a(n+1)-an={a1+n[a(n+1)-an]}-{a1+(n-1)[an-a(n-1)]}=n*a(n+1)-(2n-1)an+(n-1)*a(n-1)
所以a(n+1)-2*an+a(n-1)=0
即a(n+1)-an=an-a(n-1)
所以{an}是等差数列
再证明：若{an}是等差数列,则Sn=n/2*(an+a1)
an=a1+(n-1)d
Sn=n*a1+n(n-1)/2*d
=n/2*a1+n/2*a1+n/2*(n-1)d
=n/2*a1+n/2*[a1+(n-1)d]
=n/2*a1+n/2*an
=n/2*(an+a1)