Question

已知f(x)=根号下4+1/x2,数列{an}的前n项和为Sn,点Pn(an,-1/an+1)(n

Answer 1

f(x) =√(4+1/x^2)
Pn(an, -1/(a(n+1) ) is on y=f(x)
a1=1, an>0
Solution:
(1)
-1/a(n+1) =√(4+1/(an)^2)
1/[a(n+1)]^2 = 4+ 1/(an)^2
1/[a(n+1)]^2 - 1/(an)^2 =4
1/(an)^2 -1/(a1)^2 =4(n-1)
1/(an)^2 = 4n-3
an = 1/√(4n-3)
(2)
Tn =b1+b2+...+bn
T(n+1)/(an)^2=Tn/(a(n+1))^2+16n^2-8n-3
b1=? bn is AP
Solution
T(n+1)/(an)^2=Tn/(a(n+1))^2+16n^2-8n-3
T(n+1)(4n-3)=Tn(4n+1)+ 16n^2-8n-3
=Tn(4n+1)+ (4n-3)(4n+1)
T(n+1)/(4n+1) - Tn/(4n-3) = 1
Tn/(4n-3) -T(n-1)/(4n-7) = 1
Tn/(4n-3) -T1 =n-1
Tn = (b1+n-1)(4n-3)
=4n^2+(4b1-7)n -3(b1-1) (1)
T(n-1)=4(n-1)^2 +(4b-7)(n-1) - 3(b1-1) (2)
(1)-(2)
bn = 8n-4+(4b1-1)
b1 can be any number