■パスカルの三角形の概3等分(その9)

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

であったが,

  (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

 一般に

  (n,k)+(n,m+k)+(n,2m+k)+・・・=1/m・Σ(2cosjπ/m)^n・cos(j(n−2k)π/m)

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  cos(nπ/3)+cos((n−2)π/3)+cos((n−4)π/3)=0

より,

  (n,0)+(n,1)+(n,2)+・・・(n,n)=2^n

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