Question

这两道题怎么做！！！！

Answer 1

13.

(1)

f(x)=2sinωxcosωx+cos2ωx

=sin2ωx+cos2ωx

=√2sin(2ωx+π/4)

最小正周期T=2π/(2ω)=π/ω

又已知最小正周期为π，因此π/ω=π

ω=1

(2)

f(x)=√2sin(2x+π/4)

0≤x≤π/2，则π/4≤2x+π/4≤5π/4

-√2/2≤sin(2x+π/4)≤1

-1≤√2sin(2x+π/4)≤√2

函数的值域为[-1，√2]

14.

(1)

数列是等差数列

S3=3a2=0，a2=0

S5=5a3=-5，a3=-1

d=a3-a2=-1-0=-1

a1=a2-d=0-(-1)=1

an=a1+(n-1)d=1+(-1)(n-1)=-n+2

数列{an}的通项公式为an=-n+2

(2)

1/[a(2n-1)a(2n+1)]

=1/[-(2n-1)+2][-(2n+1)+2]

=½[-(2n-1)+2+(2n+1)-2]/[-(2n-1)+2][-(2n+1)+2]

=½[1/a(2n+1) - 1/a(2n-1)]

Tn=1/(a1a3)+ 1/(a3a5)+...+1/[a(2n-1)a(2n+1)]

=½[1/a3 -1/a1+ 1/a5 -1/a3+...+ 1/a(2n+1) -1/(2n-1)]

=½[1/a(2n+1) -1/a1]

=½[1/(-(2n+1)+2) -1/1]

=-n/(2n-1)