解：
令
an=2^n/(2^n+1)(2^n+2) = x/(2^n+1) + y/(2^n + 2) = [2^n(x+y) + (2x+y)] / (2^n+1)(2^n+2)
∴x+y=1, 2x+y=0，x=-1, y=2
∴an=1/(2^(n-1) + 1) - 1/(2^n+1)
∴sn = a1 + a2 + ... + an
= 1/2 - 1/(2^1 +1) + 1/(2^1 +1) -1/(2^2 +1) + ... + 1/(2^(n-1) + 1) - 1/(2^n+1)
= 1/2 - 1/(2^n+1)
=(2^n-1)/(2^(n+1)+2)
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