■2^n/3(その7)

 式が出そろったので,検算してみたい.

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(n,0)+(n,3)+(n,6)+・・・=(2^n+2cosnπ/3)/3

(n,1)+(n,4)+(n,7)+・・・=(2^n+2cos(n−2)π/3)/3

(n,2)+(n,5)+(n,8)+・・・=(2^n+2cos(n−4)π/3)/3

n=10とする.

(10,0)+(10,3)+(10,6)+(10,9)=1+120+210+10=341

(2^10+2cos10π/3)/3=(1024−1)/3=341

(10,1)+(10,4)+(10,7)+(10,10)=10+210+120+1=341

(2^10+2cos8π/3)/3=(1024−1)/3=341

(10,2)+(10,5)+(10,8)=45+252+45=342

(2^10+2cos6π/3)/3=(1024+2)/3=342

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(n,0)+(n,4)+(n,8)+・・・=(2^n+2^n/2・2cosnπ/4)/4

(n,1)+(n,5)+(n,9)+・・・=(2^n+2^n/2・2cos(n−2)π/4)/4

(n,2)+(n,6)+(n,10)+・・・=(2^n+2^n/2・2cos(n−4)π/4)/4

(n,3)+(n,7)+(n,11)+・・・=(2^n+2^n/2・2cos(n−6)π/4)/4

n=10とする.

(10,0)+(10,4)+(10,8)=1+210+45=256

(2^10+2^52cos10π/4)/4=(1024)/4=256

(10,1)+(10,5)+(10,9)=10+252+10=272

(2^10+2^5・2cos8π/4)/3=(1024+64)/4=272

(10,2)+(10,6)+(10,10)=45+210+1=256

(2^10+2^5・2cos6π/4)/3=(1024)/4=256

(10,3)+(10,7)=120+120=240

(2^10+2^5・2cos4π/4)/3=(1024+64)/4=272

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