Question

求解数列问题

第二问！
a1=1,an+1=√an^2-2an+2 +b
b=-1问：是否存在实数c使得a2n＜c＜a2n+1对所有n∈N*成立、

Answer 1

a(1) = 1, a(n+1) = {[a(n)]^2 - 2a(n) + 2}^(1/2) + b.
(1), b=1,

a(n+1) - 1 = {[a(n)]^2 - 2a(n) + 2}^(1/2) >= 0. a(n)>=1.

[a(n+1)-1]^2 = [a(n)]^2 - 2a(n) + 2 = [a(n)-1]^2 + 1,
{[a(n)-1]^2}是首项为[a(1)-1]^2 = 0, 公差为1的等差数列。

[a(n)-1]^2 = 0 + (n-1),
a(n) - 1 = (n-1)^(1/2),

a(n) = 1 + (n-1)^(1/2).

(2) b = -1.
a(n+1) = {[a(n)]^2 - 2a(n) + 2}^(1/2) - 1 = {[a(n)-1]^2 + 1}^(1/2) - 1 >= 1^(1/2) - 1 = 0.

a(2n+1) = {[a(2n)-1]^2 + 1}^(1/2) - 1,

a(2n+1) - a(2n) = {[a(2n)-1]^2 + 1}^(1/2) - 1 - a(2n)
= {[a(2n)-1]^2 + 1 - [1+a(2n)]^2}/ [{[a(2n)-1]^2 + 1}^(1/2) + 1 + a(2n)]
= [1 -4a(2n)] / [{[a(2n)-1]^2 + 1}^(1/2) + 1 + a(2n)]

当0<=a(2n) < 1/4时，a(2n+1) - a(2n) > 0.
此时，a(2n+1) = {[a(2n)-1]^2 + 1}^(1/2) - 1,

0<=a(2n)<1/4时，[a(2n)-1]^2单调递减。1 >= [a(2n)-1]^2 > 9/16
a(2n+1) > { 9/16 + 1}^(1/2) - 1 = {25/16}^(1/2) - 1 = 5/4 - 1 = 1/4.

而，a(2) = {[a(1)-1]^2 + 1}^(1/2) - 1 = 0, 0<= a(2) < 1/4,满足要求。
所以，存在实数c = 1/4,使得总有，a(2n) < c = 1/4 < a(2n+1).