sn=1/8(an+2)已知Sn=1/8(an+2)^2 证：{an}是等差数列

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sn=1/8(an+2)已知Sn=1/8(an+2)^2 证：{an}是等差数列
为什么这道题可以不用验n=1时是否满足啊

$sn=1/8(an+2)已知Sn=1/8(an+2)^2 证：{an}是等差数列$

Sn=1/8(an+2)^2
S(n-1)=1/8[a(n-1)+2]^2
an=Sn-S(n-1)=1/8(an+2)^2-1/8[a(n-1)+2]^2
化简得[an-a(n-1)-4][an+a(n-1)]=0,即
an-a(n-1)-4=0或an+a(n-1)=0,得
an=a(n-1)+4或an=-a(n-1)
又s1=1/8(a1+2)^2,得a1=2
第一种情况：由an=a(n-1)+4得an=a1+(n-1)d=4n-2,因此{an}是a1为2,公差为4的等差数列.
第二种情况：an=-2(-1)^n,不是等差数列.

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