# Sum with irrational powers and binomial coefficients

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.

• Well, what do you know about the binomial expansion theorem? – Simply Beautiful Art Feb 18 '17 at 0:46
• 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 – Sum'math Feb 18 '17 at 1:05
• Using that, you can then derive the closed form of your sum as martycohen has laid out for you. – Simply Beautiful Art Feb 18 '17 at 1:06
• Ah! I have worked it out now on my own – Sum'math Feb 18 '17 at 1:31
• Wonderful! And welcome to the site :D – Simply Beautiful Art Feb 18 '17 at 1:35

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

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.

• In simplest form, the answer appears to be $\frac{5}{4}F_{3n}$. How neat. :) – Sum'math Feb 18 '17 at 1:35