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Prove that \begin{eqnarray}\sum_{i=0}^k5^{k-i-1}\binom{k}{i}\left(\frac{(37+62\cos\frac{2\pi}{5}+56\cos\frac{4\pi}{5})^i}{\left(2\sin\frac{\pi}{5}\right)^{2(k-i-1)}}+\frac{(37+62\cos\frac{4\pi}{5}+56\cos\frac{8\pi}{5})^i}{\left(2\sin\frac{2\pi}{5}\right)^{2(k-i-1)}}\right)\end{eqnarray} is an integer. Can we further simplify the above expression? Thanks.

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Using the same approach as @drhab in his answer, I arrived to a slightly different result $$\color{blue}{t_k=\frac{b}{5}\left(a+\frac 5b\right)^{k}+\frac{d}{5}\left(c+\frac5d\right)^{k}}$$ Now, using the values of the trigonometric terms, we have $$a=\frac{3}{2} \left(5+\sqrt{5}\right)\qquad b=\frac{1}{2} \left(5-\sqrt{5}\right)\qquad c=\frac{3}{2} \left(5-\sqrt{5}\right)\qquad d=\frac{1}{2} \left(5+\sqrt{5}\right)$$ $$a+\frac 5b=2 \left(5+\sqrt{5}\right)\qquad c+\frac5d=2 \left(5-\sqrt{5}\right)$$ leading the the sequence $$\{1,8,80,960,12800,179200,2560000,36864000,532480000,7700480000,111411200000\}$$

We could make $t_k$ more fancy in terms of $\phi$, the golden ratio.

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  • $\begingroup$ +1 Nice answer! Btw I am a "he". $\endgroup$ – drhab Aug 31 '18 at 6:22
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You could start with something like:

$$\sum_{i=0}^{k}5^{k-i-1}\binom{k}{i}\left[\frac{a^{i}}{b^{k-i-1}}+\frac{c^{i}}{d^{k-i-1}}\right]=\frac{b}{5}\sum_{i=0}^{k}\binom{k}{i}a^{i}\left(\frac{5}{b}\right)^{k-i}+\frac{d}{5}\sum_{i=0}^{k}\binom{k}{i}c^{i}\left(\frac{5}{d}\right)^{k-i}=$$$$\frac{b}{5}\left(a+\frac{5}{b}\right)^{k}+\frac{d}{5}\left(c+\frac{5}{d}\right)^{k}$$

This to get rid of the sum.

I hope it helps, but too risky to give any guarantees :-).

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  • $\begingroup$ Using the values of $a,b,c,d$, this leads to an interesting sequence. $\endgroup$ – Claude Leibovici Aug 30 '18 at 8:28
  • $\begingroup$ @ClaudeLeibovici Why is it interesting? $\endgroup$ – Alex Fok Aug 30 '18 at 13:14
  • $\begingroup$ @AlexFok. Replace $a,b,c,d$ by their values (they are simple). Now, using the expression given in drhab's answer, compute for $k=1,2,3,4,\cdots,n$. Look at the numbers (all integers) and try to see a pattern (if any). $\endgroup$ – Claude Leibovici Aug 30 '18 at 14:54
  • $\begingroup$ I suppose that there are typo's in the answer. $\endgroup$ – Claude Leibovici Aug 31 '18 at 4:30
  • $\begingroup$ @ClaudeLeibovici thank you. Repaired. $\endgroup$ – drhab Aug 31 '18 at 6:19

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