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,a

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,a

正数列{an}的前n项之和是bnbn = a1 + a2 + …… + ancn = b1 * b2 * b3* …… bn令 n =1 b1 = a1c1 = b1因为 c1 + b1 = 1所以 a1 = b1 = c1 = 1/2b1 = a1b2 = a1 + a2c1 = b1 = a1c2 = b1*b2 = a1*(a1 + a2)b2 + c2 = 1a1 + a2 + a1*(a1 + a2) = 11/2 + a2 + (1/2)(1/2 + a2) = 1a2 = 1/6b2 = a1 + a2 = 1/2 + 1/6 = 2/3c2 = b1 * b2 = (1/2)*(2/3) = 1/3b3 = a1 + a2 + a3 = b2 + a3 = 2/3 + a3c3 = b1*b2*b3 = (1/2)(2/3)(2/3 + a3) = (1/3)(2/3 + a3) b3 + c3 = 12/3 + a3 + (1/3)(2/3 + a3) = 1a3 = 3/4 - 2/3 = 1/12b3 = a1 + a2 + a3 = 1/2 + 1/6 + 1/12 = 3/4c3 = 1 - b3 = 1/4b4 = b3 + a4 = 3/4 + a4c4 = b1 *b2*b3*b4 = (1/2)(2/3)(3/4)(3/4 + a4) = (1/4)(3/4 + a4)b4 + c4 = 13/4 + a4 + (1/4)(3/4 + a4) =1a4 = 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Ss = 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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