高一数列难题已知f（x）=-根号下4+1/x^2,点Pn（an,-1/a(n+1))在y=fx上a1=1an大于0（1）求an的通向公式（2）数列bn的前n项和Tn满足Tn+1/an^2=Tn/an^2+1+16n^2-8n-3当b1取何值时使得bn是等差数列

Question

高一数列难题
已知f（x）=-根号下4+1/x^2,点Pn（an,-1/a(n+1))在y=fx上 a1=1 an大于0 （1）求an的通向公式（2）数列bn的前n项和Tn满足Tn+1/an^2=Tn/an^2+1+16n^2-8n-3当b1取何值时使得bn是等差数列

Answer 1

(1) 令1/an=bn,则,b1=1/a1=1;b(n+1)=1/a(n+1); [b(n+1)]^2=4+bn^2
错位相减：b2^2+b3^2+...+bn^2+b(n+1)^2=4+b1^2+4+b2^2+4+b3^2+...+4+bn^2
b(n+1)^2=4n+b1^2=4n+1
a(n+1)=1/√(4n+1);an=1/√(4n-3)
(2)
Tn(1-1/an^2)=(1-1/an^2)+(4n+1)(4n-3)
当Tn=-n(n+2)时【非唯一解,2可以是其他任意正整数,合理即可,因为是数列】,bn是等差数列
此时,T1=b1=-3
T2=b1+b2=-3+b2=-8,b2=-5
T3=T2+b3=-8+b3=-15,b3=-7
T4=T3+b4=-15+b4=-24,b4=-9
.
A=-1,由Tn=1+(4n+1)(4n-3)/(1-1/an^2)=-n(n+2)可推出an=?
此略：