数列an满足递推式(a(n+2))*an-(a(n+1))^2=(t^n)*(t-1),a1=1,a2=t,(n∈N+）t＞1,且t为常数.1,求证a(n+2)-2t*a(n+1)+t*an=0(n∈N＋）;2,求证a(n+1)﹥an≥1(n∈N+)

Question

数列 an 满足递推式(a(n+2))*an-(a(n+1))^2=(t^n)*(t-1) ,a1=1,a2=t,(n∈N+）t＞1,且t为常数.
1,求证 a(n+2)-2t*a(n+1)+t*an=0(n∈N＋）;
2,求证 a(n+1)﹥an≥1(n∈N+)

Answer 1

a(n+2)an-[a(n+1)]²=(t^n)(t-1)
a(n+1)a(n-1)-(an)²=[t^(n-1)](t-1)
相减得a(n+2)an-[a(n+1)]²-a(n+1)a(n-1)+(an)²=[t^(n-1)](t-1)²=(t-1)[a(n+1)a(n-1)-(an)²]
那么a(n+2)an-[a(n+1)]²=t[a(n+1)a(n-1)-(an)²]
an[a(n+2)+tan]=a(n+1)[ta(n-1)+a(n+1)]
[a(n+2)+tan]/a(n+1)=[a(n+1)+ta(n-1)]/an=…=(a3+ta1)/a2
a3a1-a2²=t(t-1) 得a3=2t²-t
所以[a(n+2)+tan]/a(n+1)=(2t²-t+t)/t=2t
的a(n+2)-2ta(n+1)+tan=0
用数学归纳法证明
a2=t>1
所以a2>a1
假设n=k时a(k+1)>ak成立
n=k+1时,
a(k+2)=2ta(k+1)-tak=ta(k+1)+t[a(k+1)-ak]>ta(k+1)>a(k+1)
所以a(n+1)>an成立
a(n+1)>an≥a1
a(n+1)>an≥1