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What is the value of: $$\sum_{k=1}^{n} {\left(\left(-1\right)^{n-k}{n+1 \choose k+1}\left(\frac{\sqrt5}{1+\sqrt5}\left(3+\sqrt5\right)^k-\frac{\sqrt5}{1-\sqrt5}\left(3-\sqrt5\right)^k-\frac{5}{2}\right)\right)}?$$ I am stuck because of the binomial coefficient there, because without it the sum would just be a bunch of geometric series.

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    Well, what do you know about the binomial expansion theorem? – 2017-02-18
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    I know that $(a+b)^n = \sum_{k=0}^n {\left({n\choose k}a^k b^{n-k}\right)}$, if that's what you mean – 2017-02-18
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    Using that, you can then derive the closed form of your sum as martycohen has laid out for you. – 2017-02-18
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    Ah! I have worked it out now on my own – 2017-02-18
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    Wonderful! And welcome to the site :D – 2017-02-18
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    Thanks! :D And as I commented below, the answer reduces very neatly and in simplest form appears to be $\frac{5}{4}F_{3n}$. (!) – 2017-02-18
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    Oh, wow, much nicer than anything I was expecting XD – 2017-02-18

1 Answers 1

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$\sum_{k=1}^{n} {\left(\left(-1\right)^{n-k}{n+1 \choose k+1}\left(\frac{\sqrt5}{1+\sqrt5}\left(3+\sqrt5\right)^k-\frac{\sqrt5}{1-\sqrt5}\left(3-\sqrt5\right)^k-\frac{5}{2}\right)\right)}? $

Here's a start. I'm feeling too tired right now to do more.

$\begin{array}\\ \sum_{k=1}^{n} {n+1 \choose k+1} x^k &=\sum_{k=2}^{n+1} {n+1 \choose k} x^{k-1}\\ &=\frac1{x}\sum_{k=2}^{n+1} {n+1 \choose k} x^{k}\\ &=\frac1{x}\left(-1-(n+1)x+\sum_{k=0}^{n+1} {n+1 \choose k} x^{k}\right)\\ &=\frac1{x}\left((1+x)^{n+1}-1-(n+1)x\right)\\ \end{array} $

Note that $(3-\sqrt{5})(3+\sqrt{5}) =4 $.

If $x = -(3+\sqrt{5})$,

$\begin{array}\\ \sum_{k=1}^{n} {n+1 \choose k+1}(-1)^k (3+\sqrt{5})^k &=\frac1{-(3+\sqrt{5})}\left((1-(3+\sqrt{5}))^{n+1}-1+(n+1)(3+\sqrt{5})\right)\\ &=\frac{-(3-\sqrt{5})}{4}\left((-1)^{n+1}(2+\sqrt{5})^{n+1}-1+(n+1)(3+\sqrt{5})\right)\\ \end{array} $

That's all.

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    In simplest form, the answer appears to be $\frac{5}{4}F_{3n}$. How neat. :) – 2017-02-18