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I would like to get the value of the following: $$ \frac{1}{2^{n-1}} \sum_{k \ge 0} \binom {n}{2k} 5^k. $$ This comes from the computation of the trace of certain matrix.

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$$ \sum_{k \ge 0} \binom {n}{2k} (\sqrt{5})^{2k} + \sum_{k \ge 0} \binom {n}{2k+1} (\sqrt{5})^{2k+1} = (1+\sqrt{5})^n$$

$$ \sum_{k \ge 0} \binom {n}{2k} (\sqrt{5})^{2k} - \sum_{k \ge 0} \binom {n}{2k+1} (\sqrt{5})^{2k+1} = (1-\sqrt{5})^n$$

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  • $\begingroup$ In fact, I was trying to find the value of $ [ (1+ \sqrt{5})/2 ]^n + [(1- \sqrt{5})/2]^n .$ $\endgroup$ – hkju Nov 4 '14 at 21:16
  • $\begingroup$ @hkju ah ok, so I am reversing your work...anyway, it may be interesting for others...I doubt if your expression can be further simplified... $\endgroup$ – Petite Etincelle Nov 4 '14 at 21:26
  • $\begingroup$ The sequence is A000032 or A000204 in OEIS. $\endgroup$ – hkju Nov 4 '14 at 23:10
  • $\begingroup$ @hkju Yeah, it's also related to Fibonacci numbers $\endgroup$ – Petite Etincelle Nov 4 '14 at 23:24

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