设正数列{an}的前n项之和是bn,数列{bn}的前n项之积是cn,若bn+cn=1

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设正数列{an}的前n项之和是bn,数列{bn}的前n项之积是cn,若bn+cn=1
(1)求a1,a2,a3,a4
(2)猜想数列{an}的通项公式
(3)求数列{an}的前n项和S

正数列{an}的前n项之和是bn
bn = a1 + a2 + …… + an
cn = b1 * b2 * b3* …… bn
令 n =1
b1 = a1
c1 = b1
因为 c1 + b1 = 1
所以 a1 = b1 = c1 = 1/2
b1 = a1
b2 = a1 + a2
c1 = b1 = a1
c2 = b1*b2 = a1*(a1 + a2)
b2 + c2 = 1
a1 + a2 + a1*(a1 + a2) = 1
1/2 + a2 + (1/2)(1/2 + a2) = 1
a2 = 1/6
b2 = a1 + a2 = 1/2 + 1/6 = 2/3
c2 = b1 * b2 = (1/2)*(2/3) = 1/3
b3 = a1 + a2 + a3 = b2 + a3 = 2/3 + a3
c3 = b1*b2*b3 = (1/2)(2/3)(2/3 + a3) = (1/3)(2/3 + a3)
b3 + c3 = 1
2/3 + a3 + (1/3)(2/3 + a3) = 1
a3 = 3/4 - 2/3 = 1/12
b3 = a1 + a2 + a3 = 1/2 + 1/6 + 1/12 = 3/4
c3 = 1 - b3 = 1/4
b4 = b3 + a4 = 3/4 + a4
c4 = b1 *b2*b3*b4 = (1/2)(2/3)(3/4)(3/4 + a4) = (1/4)(3/4 + a4)
b4 + c4 = 1
3/4 + a4 + (1/4)(3/4 + a4) =1
a4 = 4/5 - 3/4 = 1/20
综上所述 a1 = 1/2 ,a2 = 1/6,a3 = 1/12,a4 = 1/20
数列{an}的通项公式
an = 1/[n(n+1)] = 1/n - 1/(n+1)
数列{an}的前n项和S
s = a1 + a2 + …… + an
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 …… + 1/(n-1) - 1/n + 1/n - 1/(n+1)
= 1 - 1/(n+1)
= n/(n+1)

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