Question

高一数学，一道数列大题

Answer 1

答案如图:

追问

谢谢^ω^

追答

不用谢

Answer 2

4Sn=(2n-1)a(n+1)+1

4S(n-1)=(2n-3)an+1
4[Sn-(n-1)]=(2n-1)a(n+1)-(2n-3)an
4an=(2n-1)a(n+1)-(2n-3)an
(2n+1)an=(2n-1)a(n+1)
a(n+1)/an=(2n+1)/(2n-1)
累乘得
a(n+1)/a1=(a2/a1)*(a3/a2)*...*a(n+1)/an=3*(5/3)*(7/5)*...*(2n+1)/(2n-1)=2n+1
a(n+1)/a1=2n+1
a(n+1)=a1*(2n+1)=1*(2n+1)=2n+1
a(n+1)=2n+1
an=2n-1
2)Cn=1/an*(an+2)=1/(2n-1)(2n+1)=1/2[1/(2n-1)-1/(2n+1)]
Tn=1/2[(1-1/3)+(1/3-1/5)+...+[1/(2n-1)-1/(2n+1)]=1/2[1-1/(2n+1)]=n/(2n+1)
Tn=n/(2n+1)
3)T1=1/3
kx^2-6k+k+7+3*1/3>0
kx^2-6k+k+8>0
k>0,判别式<0
36k^2-4k^2-32k<0
k^2-k<0
0<k<1