■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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