已知数列｛an}的前n项和为Sn,且满足Sn+1-2=Sn+an+2n(n为正整数）,且S2=8.求｛an} 的通项公式
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S(n+1)-2=Sn + an + 2n.(A)
Sn -2=S(n-1)+a(n-1)+2(n-1).(B)
(A)-(B)得,a(n+1)= an + an - a(n-1) + 2
移项得,a(n+1)-an = an - a(n-1) + 2
an-a(n-1) = a(n-1) - a(n-2) + 2
a(n-1)-a(n-2) = a(n-2) - a(n-3) + 2
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a3 - a2 = a2 - a1 + 2
叠加得,a(n+1) - a2 = an - a1 + 2(n-1)
a(n+1) - an = a2 - a1 + 2(n-1)
设n=1 S2 - 2 = S1 + a1 + 2 a1=2 a2=6
a(n+1) - an = 4 + 2(n-1)
an - a(n-1) = 4 + 2(n-2)
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a3 - a2 = 4 + 2
a2 - a1 = 4
a(n+1) - a1 = 4n + 2* (n-1)n/2 = n^2 + 3n
a(n+1) = n^2 + 3n + 2 = (n+1)(n+2)
an = n(n+1)
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