1)
an = 1/2*(√Sn + √S(n-1)）
而:an = Sn-S(n-1) = [√Sn + √S(n-1)][√Sn - √[S(n-1)]
所以:√Sn - √[S(n-1)] = 1/2
叠代加和,得:√Sn - √S1 = (n-1)/2
即:√Sn = (n-1)/2 + √a1
由:S4=4,得:a1 = 1/4
所以:
Sn = n^2/4
(2)
an = Sn - S(n-1) = n^2/4 - (n-1)^2/4 = (2n-1)/4