Question

数学数列中档题：正数数列｛An｝和｛Bn｝满足：对任意的正整数n，An，Bn，An+1成等差数列，Bn，An+……

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Answer 1

(1)an,bn,a(n+1)成等差数列
2bn = an+ a(n+1)
bn,a(n+1),b(n+1)成等比数列
[a(n+1)]^2 = bn.b(n+1)
√b(n+1) - √bn

= a(n+1)/√bn - √bn
= (2bn-an)/√bn - √bn
= √bn - an/√bn
=√bn - √[b(n-1).bn]/√bn
=√bn - √b(n-1)
=>{√bn}是等差数列
(2)(a2)^2 = b1b2,得b2=9/2,√b2 - √b1 =√2/2=d,

√bn =√b1 +(n-1)d=√2(n+1)/2,bn=(n+1)^2/2
而an^2=bnxbn-1=(n+1)^2/2xn^2/2=n^2x(n+1)^2/4
an=n(n+1)/2.
(3)1/an=2/n(n+1),sn=2(1-1/2+1/2-1/3+.......1/(n-1)-1/n)=2(1-1/n)=2(n-1)/n。