已知无穷数列an的前n项和Sn并且an+sn=1 n在正整数范围 求的的通项公式
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1 = a(n) + s(n),

1 = a(1) + s(1) = 2a(1),a(1)=1/2,

1 = a(n+1) + s(n+1),

0 = 1 - 1 = [a(n+1)+s(n+1)] - [a(n)+s(n)] = a(n+1)-a(n) + [s(n+1)-s(n)] = 2a(n+1) - a(n),

a(n+1) = a(n)/2,

{a(n)}是首项为a(1)=1/2,公比为1/2 的等比数列.

a(n) = (1/2)(1/2)^(n-1) = (1/2)^n = 1/2^n

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