求助一道高中数列的题

设a(n+1)-x=(√2-1)(an-x)
x-(√2-1)x=2*(√2-1)
x=√2
cn=an-√2
c(n+1)=(√2-1)cn
c1=2-√2
cn=√2*(√2-1)^n
an=cn+√2=√2((√2-1)^n+1)
2、
有点复杂
用数学归纳法证明bn>√2
b1=2>√2
假设bk>√2
b(k+1)
=(3bk+4)/(2bk+3)
=3/2-1/2(2bk+3)
>3/2-1/2(2√2+3)
=3/2-(3-2√2)/2
=√2
即b(k+1)>√2
故bn>√2

a(4n-3)=√2((√2-1)^(4n-3)+1)
a(4-3)=a1=2
b1<=a(4-3)成立

假设bk<=a(4k-3)
即bk<=√2((√2-1)^(4k-3)+1)

2b(k+1)
=3-1/(2bk+3)
<=3-1/(2*√2((√2-1)^(4k-3)+1)+3)
=3-1/(2*√2(√2-1)^(4k-3)+2*√2+3)

=3-(3+2*√2-2*√2(√2-1)^(4k-3))/((2*√2+3)²-(2*√2(√2-1)^(4k-3))²)
设分母=t

原式=3-（3+2*√2）/t+2*√2(√2-1)^(4k-3))/t

t=(2*√2+3)²-(2*√2(√2-1)^(4k-3))²
=((2*√2+3)²-8((√2-1)^(4k-3))²)

易知0<8((√2-1)^(4k-3))²)<8
((2*√2+3)²>9

故1<t<(2*√2+3)²=(√2+1)^4

故