设数列满足函数sn=an^2/4+n,an>0求an表达式

Question

设数列满足函数sn=an^2/4+n,an>0 求an表达式

Answer 1

LS的……那方法你能做出来你就神了……
Sn = an^2/4 + n
Sn-1 = an-1^2/4 + n - 1 ←注意,这里等号右面第一项是an-1而不是an,LS就错在这里
an = Sn - Sn-1 = (an^2 - an-1^2)/4 +1
an^2 - 4an + 4 = an-1^2
(an - 2)^2 = an-1^2
|an -2| = |an-1|
又因为S1 = a1,a1 = a1^2/4 + 1,a1 = 2
已知了an>0,绝对值可以去掉,an - 2 = an-1
所以an = an-1 + 2 = an-2 + 4 = ...= a1 + 2n-2 = 2n