数学红对勾例2：若数列{an}满足a1=1,且an+1=an/(1+an),设数列{bn}的前n项和为Sn,且Sn=2-
1个回答
a(n+1)=an/(1+an)
1/a(n+1) = (1+an)/an
= 1+1/an
1/a(n+1) -1/an =1
1/an -1/a1=n-1
an = 1/n
Sn=2-bn
n=1, b1= 1
bn = Sn -S(n-1)
=-bn+b(n-1)
bn = (1/2)b(n-1)
= (1/2)^(n-1) .b1
=(1/2)^(n-1)
let
S=1.(1/2)^0+2.(1/2)^1+...+n.(1/2)^(n-1) (1)
(1/2)S= 1.(1/2)^1+2.(1/2)^2+...+n.(1/2)^n (2)
(1)-(2)
(1/2)S= (1+1/2+...+1/2^(n-1)) -n.(1/2)^n
= 2(1-(1/2)^n)-n.(1/2)^n
S = 4(1-(1/2)^n)-2n.(1/2)^n
=4-(2n+4)(1/2)^n
cn =bn/an
=n.(1/2)^(n-1)
c1+c2+...+cn = S =4-(2n+4)(1/2)^n
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