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Question:

Find this value

$$\sum_{i_{1} = 1}^{n}\sum_{i_{2} = 1}^{n}\cdots\sum_{i_{k} = 1}^{n} \cos\left(k\left[\vphantom{\Large A}\, i_{1}^{k} + i_{2}^{k} + \cdots +i_{k}^{k}\,\right] \over n\right) $$

This is an interesting problem,

My try: Let $$x=\sum_{i_1=1}^{n}\sum_{i_2=1}^{n}\cdots\sum_{i_k=1}^{n}\cos\frac{k(i^k_1+i^k_2+\cdots+i^k_k)}{n}$$ $$y=\sum_{i_1=1}^n \sum_{i_2=1}^n \cdots\sum_{i_k=1}^n \sin \frac{k(i^k_1+i^k_2 + \cdots+i^k_k)}{n}$$ so $$x+y=\cdots=f(n)?$$ $$xy=\cdots=g(n)?$$ But at last I failed.

This problem is from a magazine's open problem (2010). Thank you for your help.

Michael Kirchner
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math110
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  • In other words $$x=\sum_{i_1=1}^{n}\sum_{i_2=1}^{n}\cdots\sum_{i_k=1}^{n}\cos{\frac kn\displaystyle\sum_{j=1}^k i^k_j}$$ – Ali Caglayan Nov 30 '13 at 17:13
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    The only simplification I can think of is $$x + iy = \sum_{m_1=1}^n \cdots \sum_{m_k=1}^n e^{i \frac{k}{n} (m_1^k + \cdots m_k^k)} = \sum_{m_1=1}^n e^{i \frac{k}{n} m_1^k} \times \cdots \times \sum_{m_k=1}^n e^{i \frac{k}{n} m_k^k} = \left(\sum_{m=1}^n e^{i \frac{k}{n} m^k}\right)^k $$ No idea how to make this any further. – achille hui Dec 01 '13 at 05:16
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    Is it $\cos\frac{k(i^k_1+i^k_2+\cdots+i^k_k)}{n}$ or $\cos\frac{k(i^k_1+i^k_2+\cdots+i^k_k)\color{red}{\pi}}{n}$? – Shane Dec 09 '13 at 14:01

0 Answers0