據(jù)

f(x+1)=-1/2f(x)+√3/2g(x)

g(x+1)=-1/2g(x)-√3/2f(x)

有

g(x+1)

=

-1/2(2√3/3f(x+1)+√3/3f(x)）

-√3/2f(x)

=

-√3/3f(x+1)-√3/6f(x)-√3/2f(x)

=

-√3/3f(x+1)-2√3/3f(x)

=

-√3/3(-1/2f(x)+√3/2g(x))

-2√3/3f(x)

=

√3/6f(x)-1/2g(x)-2√3/3f(x)

=

-√3/2f(x)-1/2g(x)

即

g(x+2)

=

-1/2g(x+1)-√3/2f(x+1)

=

-1/2(-√3/2f(x)-1/2g(x))

-√3/2(-1/2f(x)+√3/2g(x))

=

√3/4f(x)+1/4g(x)

+√3/4f(x)-3/4g(x)

=

√3/2f(x)-1/2g(x)

即

f(x+2)

=

-1/2f(x+1)+√3/2g(x+1)

=

-1/2(-1/2f(x)+√3/2g(x))

+√3/2(-1/2g(x)-√3/2f(x))

=

1/4f(x)-√3/4g(x)

-√3/4g(x)-3/4f(x)

=

-1/2f(x)-√3/2g(x)

即

f(x+3)

=

-1/2f(x+2)+√3/2g(x+2)

=

-1/2(-1/2f(x)-√3/2g(x))

+√3/2(√3/2f(x)-1/2g(x))

=

1/4f(x)+√3/4g(x)

+3/4f(x)-√3/4g(x)

=

f(x)

有

f(x+2)

=

-1/2f(x+1)+√3/2g(x+1)

=

1/4f(x)-√3/4g(x)

+√3/2(-1/2g(x)-√3/2f(x))

=

1/4f(x)-√3/4g(x)

-√3/4g(x)-3/4f(x)

=

-1/2f(x)-√3/2g(x)

且

g(x+2)

=

-1/2g(x+1)-√3/2f(x+1)

=

1/4g(x)+√3/4f(x)

-√3/2(-1/2f(x)+√3/2g(x))

=

1/4g(x)+√3/4f(x)

+√3/4f(x)-3/4g(x)

=

√3/2f(x)-1/2g(x)

即

g(x+3)

=

-1/2g(x+2)-√3/2f(x+2)

=

-√3/4f(x)+1/4g(x)

-√3/2(-1/2f(x)-√3/2g(x))

=

-√3/4f(x)+1/4g(x)

+√3/4f(x)+3/4g(x)

=

g(x)

即

g(2)=g(365)=-√3

據(jù)

f(x)=f(5-x)

設(shè)

x=2

有

f(2)=f(3)

且

f(3)=-1/2f(2)+√3/2g(2)

即

3/2f(2)=√3/2g(2)

即

√3f(2)=g(2)=-√3

即

f(2)=-1

即

f(3)=-1

有

g(x+1)

=

-√3/3f(x+1)-2√3/3f(x)

設(shè)

x=1

有

g(2)=-√3/3f(2)-2√3/3f(1)

即

-√3=√3/3-2√3/3f(1)

即

-4√3/3=-2√3/3f(1)

即

f(1)=2

即

Σ(k=1，2023)f(k)

=

674(f(1)+f(2)+f(3))+f(1)

=

674(2-1-1)+2

=

2

ps.

特殊

求解

除

視頻

擬定設(shè)法

抑或

設(shè)

f(x)=-2sin(2π/3x-7π/6)

g(x)=-2cos(2π/3x-7π/6)

求解過(guò)程如下

設(shè)

f(x)=Asin(ωx+φ)

g(x)=Acos(ωx+φ)

有

f(x+1)

=

Asin(ω(x+1)+φ)

=

Asin(ωx+ω+φ)

且

f(x+1)

=

-1/2f(x)+√3/2g(x)

=

cos(2π/3)Asin(ωx+φ)

+sin(2π/3)Acos(ωx+φ)

=

Asin(ωx+2π/3+φ)

即

ω=2π/3

據(jù)

f(x)=f(5-x)

設(shè)

x=5/2

2π/3x+φ=π/2

即

5π/3x+φ=π/2

即

φ=-7π/6

據(jù)

g(365)=-√3

有

Acos(730π/3-7π/6)

=

Acos(243π+π/3-7π/6)

=

Acos(243π-5π/6）

=

-Acos5π/6

=

√3A/2

=

-√3

即

A=-2

即

f(x)=-2sin(2π/3x-7π/6)

有

T=2π/(2π/3)=3

f(1)

=-2sin(2π/3-7π/6)

=2

f(2)

=-2sin(4π/3-7π/6)

=-1

f(3)

=-2sin(2π-7π/6)

=-1

即

Σ(k=1，2023)f(k)

=

674(f(1)+f(2)+f(3))+f(1)

=

674(2-1-1)+2

=

2

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