Question

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

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 0.

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

0 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