Question

设各项均为正数的数列｛an}和｛bn}满足：an,bn,an+1成等差数列，bn,an+1,bn+

设各项均为正数的数列｛an}和｛bn}满足：an,bn,an+1成等差数列，bn,an+1,bn+1成等比数列，且a1=1,b1=2,a2=3,求通项an,bn

Answer 1

an,bn,a(n+1)成等差数列 =>
an+a(n+1)=2bn (1)

bn,a(n+1),b(n+1)成等比数列 =>

bn.b(n+1) = [a(n+1)]^2 (2)

sub(2) into (1)
√[b(n-1).bn]+√[bn.b(n+1)]=2bn
√b(n-1)+√b(n+1)=2√bn
=> √bn 成等差数列

a1=1,b1=2,a2=3

from (2)
b1.b2 = (a2)^2
2b2 = 9
b2 =9/2
d = √b2-√b1 = 3√2/2 - √2 = √2/2

√bn=√b1+(n-1)d
=√2 + (n-1)√2/2
= (n+1)√2/2
bn =(n+1)^2/2

from (1)
an+a(n+1)=2bn
=(n+1)^2
let
a(n+1) +k1(n+1)^2 +k2(n+1)+k3 = -[an + k1n^2+k2n+k3]

coef. of n^2
-2k1 = 1
k1 = -1/2
coef.of n
-2k2-2k1=2
1-2k2=2
k2 =-1/2
coef. of constant
-2k3-k1-k2 =1
-2k3+1/2+1/2 =1
k3 =0

an+a(n+1)=(n+1)^2
a(n+1) -(1/2)(n+1)^2 -(1/2)(n+1) = -[an - (1/2)n^2-(1/2)n]
let
cn =an - (1/2)n^2-(1/2)n
cn 是等比数列, q= -1
a(n+1) -(1/2)(n+1)^2 -(1/2)(n+1) = -[an - (1/2)n^2-(1/2)n]
(an - (1/2)n^2-(1/2)n) = (-1)^(n-2) . [a2 - (1/2)2^2-(1/2)2]
= (-1)^(n-2) . (3-2-1)
= 0
an = (1/2)n^2+(1/2)n

追问

有另一种方法吗？

追答

from (2)
bn.b(n+1) = [a(n+1)]^2

[a(n+1)]^2 = [(n+1)^2/2](n+2)^2/2]
a(n+1) = (n+1)(n+2)/2
an= n(n+1)/2