$\cos(π(2j^2-n)/4n)$求和

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$$\sqrt{2n}=\sum _{j=0}^{2 n-1} \cos \left(\frac{\pi  \left(2 j^2-n\right)}{4 n}\right)$$与$$0=\sum _{j=0}^{2 n-1} \sin\left(\frac{\pi  \left(2 j^2-n\right)}{4 n}\right)$$

hbghlyj

顶啊✊