题目

将tn-1*tn+1=tn*tn+5转换成为递推式,已知t1=1,t2=2
注意：其中n-1,n+1,n为下标

提问时间：2021-01-03

答案

t(n-1)*t(n+1)=tn*tn+5
当n=2时,t1*t3=(t2)^2+5,t3=9
当n=3时,t2*t4=(t3)^2+5,t4=43
(tn)^2-t(n-1)t(n+1)+5=0
[t(n-1)]^2-t(n-2)tn+5=0
两式相减:
(tn)^2-[t(n-1)]^2=t(n-1)t(n+1)-t(n-2)tn
(tn)^2+t(n-2)tn=[t(n-1)]^2+t(n-1)t(n+1)
[tn+(1/2)t(n-2)]^2=[t(n-1)+(1/2)t(n+1)]^2
tn+(1/2)t(n-2)=±[t(n-1)+(1/2)t(n+1)]
t(n+2)+(1/2)tn=±[t(n+1)+(1/2)t(n+3)]
1.
取+时：t(n+2)+(1/2)tn=t(n+1)+(1/2)t(n+3)
t(n+2)-t(n+1)=(1/2)t(n+3)-(1/2)tn
=(1/2)[t(n+3)-tn]
=(1/2)[t(n+3)-t(n+2)+t(n+2)-t(n+1)+t(n+1)-tn]
=(1/2)[t(n+3)-t(n+2)]+(1/2)[t(n+2)-t(n+1)]+(1/2)[t(n+1)-tn]
设an=t(n+1)-tn,a1=t2-t1=1,a2=t3-t2=7
a(n+1)=(1/2)a(n+2)+(1/2)a(n+1)+(1/2)an
a(n+1)=a(n+2)+an
a(n+2)-a(n+1)+an=0
设a(n+2)-xa(n+1)=y[a(n+1)-xan]
x+y=1,xy=1
x1=(1+√5)/2,y1=(1-√5)/2
x2=(1-√5)/2,y2=(1+√5)/2
a(n+2)-xa(n+1)=[a2-xa1]y^n=(7-x)y^n
设[a(n+2)+z]-x[a(n+1)+z]=0
a(n+2)-xa(n+1)=xz-z
z=[(7-x)y^n]/(x-1)
[a(n+2)+z]=x[a(n+1)+z]
a(n+2)+z=(a1+z)x^(n-1)
a(n+2)=-z+(a1+z)x^(n-1)
an=-z+(1+z)x^(n-3)=t(n+1)-tn
t(n+1)-tn=-z+(1+z)x^(n-3)
=-[(7-x)y^n]/(x-1)+{1+[(7-x)y^n]/(x-1)}x^(n-3)
tn-t(n-1)=
t(n-1)-t(n-2)=
……
t2-t1=
两边相加得出结果：
2.
取-时：t(n+2)+(1/2)tn=-t(n+1)-(1/2)t(n+3)