Question

在数列an中,a1=1,an+1=an²-1,则数列前n项则数列前n项和为

如题，其中n均为下表，写下过程，谢啦

Answer 1

如图所示：望采纳哦~如有疑问可以追问。

追问

不好意思，我打的不清楚。n+1是a的下标- -

追答

Answer 2

a(1)=1,
a(2) = [a(1)]^2 - 1 = 0,
a(3) = [a(2)]^2 - 1 = -1,
a(4) = [a(3)]^2 - 1 = 0,
a(5) = [a(4)]^2 - 1 = -1,
...
a(2n) = 0,
a(2n+1) = -1.

记s(n) = a(1) + a(2) + ... + a(n).
则，s(1) = a(1) = 1.
s(2) = s(1) + a(2) = s(1) = 1.
s(3) = s(2) + a(3) = 1 - 1 = 0
s(4) = s(3) + a(4) = s(3) = 0.
s(5) = s(4) + a(5) = -1.
s(6) = s(5) + a(6) = s(5) = -1.
s(7) = s(6) + a(7) = -2.
...
s(2n) = s(2n-1),
s(2n+1) = a(1) - n = 1-n.

s(1) = s(2) = 1,
s(2n+1) = 1-n,
s(2n+2) = s(2n+1) = 1-n.

综合，有
s(2n-1) = 1 - (n-1) = 2-n,
s(2n) = s(2n-1) = 2-n.