1.
-1≤cosɑ≤1
即當(dāng)-1≤x≤1時,f(x)＜0
1≤2－sinß≤3
即當(dāng)1≤x≤3時,f(x)≥0,
所以f(1)=0
f(1)=1/4＋b-3/4=b-1/2=0
b=1/2;
2.
f(x)=(1/4)x^2＋(1/2)x-3/4
Sn=f(an)=(1/4)(an)^2＋(1/2)an-3/4
S(n-1)=(1/4)[a(n-1)]^2＋(1/2)a(n-1)-3/4
兩式相減：
an=Sn-S(n-1)=(1/4)(an)^2＋(1/2)an-(1/4)[a(n-1)]^2-(1/2)a(n-1)
(an)^2-2an-[a(n-1)]^2-2a(n-1)=0
[an+a(n-1)][an-a(n-1)-2]=0
an-a(n-1)=2
an=a1+2(n-1)
又S1=f(a1)=(1/4)(a1)^2＋(1/2)a1-3/4=a1
(1/4)(a1)^2-(1/2)a1-3/4=0
(a1)^2-2a1-3=0
a1=3
所以an=a1+2(n-1)=3+2(n-1)=2n+1;
3.
Cn=1/(1+an)^2=1/(2n+2)^2=(1/4)[1/(n+1)^2]
4Cn=1/(n+1)^2
又n(n+1)＜(n+1)^2＜(n+1)(n+2)
1/[(n+1)(n+2)]＜1/(n+1)^2＜1/[n(n+1)]
1/(n+1)-1/(n+2)＜1/(n+1)^2＜1/n-1/(n+1)
1/n-1/(n+1)＜1/n^2＜1/(n-1)-1/n
1/(n-1)-1/n＜1/(n-1)^2＜1/(n-2)-1/(n-1)
……
1/3-1/4＜1/3^2＜1/2-1/3
1/2-1/3＜1/2^2＜1/1-1/2
兩邊相加：
1/2-1/(n+2)＜1/(n+1)^2+1/n^2+1/(n-1)^2+……+1/3^2+1/2^2＜1-1/(n+1)
1/2-1/(n+2)＜4Tn＜1-1/(n+1)
1/8-1/(4n+8)＜Tn＜1/4-1/(4n+4)