已知数列an,an属于n*,sn=1/8*(an+2)^2,{an}是等差数列_数学解答

已知数列an,an属于n*,sn=1/8*(an+2)^2,{an}是等差数列

人气：482 ℃　时间：2020-06-27 09:49:56

解答

Sn=1/8*(an+2)^2,S(n-1)=1/8*[a(n-1)+2]^2.所以Sn-S(n-1)=an=1/8*[an^2-a(n-1)^2]+1/2*[an-a(n-1)],所以1/2*[an+a(n-1)]=1/8*[an^2-a(n-1)^2],即an-a(n-1)=4