Question

各项均为正数的数列{an},其前n项的和为Sn,

且Sn=(根号S(n-1 )+根号a1)^2(n≥2),若bn=a(n+1)/an+an/a(n+1)，
且数列bn前n项的和为Tn，则Tn为多少，（要详细过程）

Answer 1

Sn=(√S(n-1)+√a1)^2
=S(n-1)+2√a1S(n-1)+a1^2
Sn-S(n-1)=2√a1S(n-1)+a1^2=an
2√a1S(n-1)=an-a1^2

4a1S(n-1)=an^2-2a1an+a1^4
4a1S(n-2)=a(n-1)^2-2a1a(n-1)+a1^4
4a1[S(n-1)-S(n-2)]=an^2-a(n-1)^2-2a1[an-a(n-1)]
4a1a(n-1)=an^2-a(n-1)^2-2a1an+2a1a(n-1)
an^2-a(n-1)^2-2a1an-2a1a(n-1)=0

[an+a(n+1)][an-a(n-1)]-2a1[an+a(n+1)]=0
[an+a(n+1)][an-a(n-1)-2a1]=0
∵各项均为正数
∴an-a(n-1)-2a1=0
an-a(n-1)=2a1
a(n-1)-a(n-2)=2a1
..............
a2-a1=2a1
an-a1=2(a1+a1+.......a1)
=2(n-1)a1
=2a1n-2a1
an=2a1n-a1
=a1(2n-1)

bn=a(n+1)/an+an/a(n+1)
=a1(2(n+1)-1)/a1(2n-1) +a1(2n-1)/a1(2(n+1)-1)
=(2n+1)/(2n-1)+(2n-1)/(2n+1)
=(4n^2+4n+1+4n^2-4n+1)/(4n^2-1)
=(8n^2+2)/(4n^2-1)
=(8n^2-2+4)/(4n^2-1)
=2+4/(2n+1)(2n-1)
=2[1+1/(2n-1)-1/(2n+1)]
Tn=b1+b2+.............bn
=2[1+1-1/3+1+1/3-1/5+...........+1+1/(2n-1)-1/(2n+1)]
=2[n+1-1/(2n+1)]
=2[n+(2n+1-1)/(2n+1)]
=2[n+2n/(2n+1)]
=(4n^2+6n)/(2n+1)

Answer 2

因为Sn=(根号S(n-1 )+根号a1)^2(
把n=2代入得a1+a2=4a1
则a2=3a1
把n=3代入得a3=5a1
……
sn=s(n-1)+a1+2根号S(n-1 )*根号a1
因此，an=sn-s(n-1)=a1+2根号S(n-1 )*根号a1
而a(n+1)=3a1+2根号S(n-1 )*根号a1
即a(n+1)=an+2a1
由此可知an=(2n-1)a1
sn=n^2a1

bn=a(n+1)/an+an/a(n+1)=(2n+1)/(2n-1)+(2n-1)/(2n+1)
数列bn前n项的和为Tn=(3/1+1/3)+(5/3+3/5)+……+[(2n-1)/(2n-3)+(2n-3)/(2n-1)]+[(2n+1)/(2n-1)+(2n-1)/(2n+1)]
=3+2(n-1)+(2n-1)/(2n+1)
=(2n+1)+(2n-1)/(2n+1)

Answer 3

这个不是已经做过了么？