高中数学题目,高手请进nC0+nC3+nC6+nC9+.=(2^n+2cos(nπ/3))/3 求证这个式子C指的就是排

我把组合数记作C(n,0)等等．ω＝cos(2π/3)+isin(2π/3),
ω＾2＝cos(4π/3)+isin(4π/3),ω^3＝cos(6π/3)+isin(6π/3)=1.
1+ω+ω^2=0.
(1+x)^n=C(n,0)+C(n,1)x+C(n,2)x^2+C(n,3)x^3+.
令x=1,得C(n,0)+C(n,1)+C(n,2)+C(n,3)+.=2^n.(1)
令x=ω,得C(n,0)+C(n,1)ω+C(n,2)ω^2+C(n,3)ω^3+.=(1+ω)^n
=[cos(π/3)+isin(π/3)]^2=cos(nπ/3)+isin(nπ/3).(2)
令x=ω^2,得C(n,0)+C(n,1)ω^2+C(n,2)ω^4+C(n,3)ω^6+...=(1+ω^2)^2
=[cos(π/3)-isin(π/3)]^2=cos(nπ/3)-isin(nπ/3).(3)
(1),(2),(3)三式相加,得
3[C(n,0)+C(n,3)+.]=2^n+2cos(nπ/3).