Question

已知数列{an}的前n项和为sn,且4sn=an²+2an,a1+a2≠0,a1≠0。（1）求证

数列{an}是等差数列

Answer 1

4s1=a1²+2a1
即4a1=a1²+2a1
得：a1=2

Sn=1/4[an²+2an]
an=Sn-S(n-1)=1/4[an²+2an-a(n-1)²-2a(n-1)]
化得：an²-2an-a(n-1)²-2a(n-1)=0
[an+a(n-1)][an-a(n-1)]-2[an+a(n-1)]=0
[an+a(n-1)[an-a(n-1)-2]=0
因为a1+a2≠0，故有an-a(n-1)-2=0
即{an}是公差为2的等差数列。