$\exp((j+k)(j-k+1)\frac{iπ}{2n})$求和

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$$n=\sum _{j=0}^{2 n-1} \sum _{k=0}^{2 n-1-j} \exp \left(\frac{i \pi  }{2 n}(j+k) (j-k+1)\right)$$

In[]:= Table[RootReduce@Sum[Exp[I Pi/(2n)(k+j)(j-k+1)],{j,0,2n-1},{k,0,2n-1-j}],{n,10}]

Out[]= {1,2,3,4,5,6,7,8,9,10}

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🕵️列表观察规律

1. Do[Print@Grid[Table[Exp[I Pi/(2n)Mod[(k+j)(j-k+1),4n,-2n]],{j,0,2n-1},{k,0,2n-1-j}],Frame->All],{n,2,4}]

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n=2
n=3
n=4

蓝色区域沿横轴线$j=\frac{(n-1)+n}2$消掉     $\forall k\in\{0,\dots,j\}:$\[0=\exp\left({i\pi\over 2n}(j+k)(j-k+1)\right)+\exp\left({i\pi\over 2n}((2n-1-j)+k)((2n-1-j)-k+1)\right)\]
红色区域沿纵轴线$k=\frac{n+(n+1)}2$消掉     $\forall k\in\{j+2,\dots,2n-1-j\}:$\[0=\exp\left({i\pi\over 2n}(j+k)(j-k+1)\right)+\exp\left({i\pi\over 2n}(j+(2n+1-k))(j-(2n+1-k)+1)\right)\]
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