数列{an}中,已知sn=an-1/sn-2,①:求出s1,s2,s3,s4,②:猜想数列{an}的前n项和sn的公式,

数列{a(n)}中,已知s(n) = a(n) - 1/s(n) - 2,①：求出s(1),s(2),s(3),s(4),②：猜想数列{a(n)}的前n项和s(n)的公式,并加以证明
s(1) = a(1) = a(1) - 1/s(1) - 2,
0 = -1/s(1) - 2,
s(1) = -1/2.
s(2) = s(1) + a(2) = -1/2 + a(2) = a(2) - 1/s(2) - 2,
-1/2 = -1/s(2) - 2,
s(2) = -2/3.
s(3) = s(2) + a(3) = -2/3 + a(3) = a(3) - 1/s(3) - 2,
-2/3 = -1/s(3) -2,
s(3) = -3/4.
s(4) = s(3) + a(4) = -3/4 + a(4) = a(4) - 1/s(4) - 2,
-3/4 = -1/s(4) -2,
s(4) = -4/5.
① s(1) = -1/2,s(2) = -2/3,s(3) = -3/4,s(4) = -4/5.
② 猜想数列{a(n)}的前n项和s(n) = -n/(n+1)
证明,
(1)n = 1,s(1) = -1/2.符合猜想.
（2）假设n = k时,有 s(k) = -k/(k+1),
则n = k+1时,有
s(k+1) = s(k) + a(k+1) = -k/(k+1) + a(k+1) = a(k+1) - 1/s(k+1) - 2,
-k/(k+1) = -1/s(k+1) - 2,
s(k+1) = -(k+1)/(k+2).
符合猜想.
因此,由归纳法证得,
数列{a(n)}的前n项和s(n) = -n/(n+1),n = 1,2,...
的结论成立.