已知数列｛an｝和｛bn｝满足a1a2···an=2^(bn-n),若｛an｝为等比数列，且a1=1,b2=b1+2.

(1)
an = a1q^(n-1) (1)
a1.a2....an = 2^(bn-n) (2)

a1=1, from (2) , n=1

a1 = 2^(b1-1)
b1=1

b2 = b1+2 = 3

from (2) , n=2
a1.a2 = 2^(b2-2)
a2 = 2

from (1)
a2/a1 = q = 2
ie an = 2^(n-1)

from (2)
a1.a2 ...an = 2^(bn-n)
2^[ n(n-1)/2] = 2^(bn-n)
bn - n = n(n-1)/2
bn = n(n+2)/2

(2)
cn = 1/an -1/bn
= 2^(1-n) - 2/[n(n+2)]
= (1/2)^(n-1) - [ 1/n - 1/(n+2) ]

Sn = c1+c2+...+cn
= (1- 1/2^n) - [ 1 +1/2 - 1/(n+1) -1/(n+2) ]
= -1/2 - (1/2)^n + { 1/[(n+1)(n+2)] }