3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Well, what do you know about the binomial expansion theorem? $\endgroup$ – Simply Beautiful Art Feb 18 '17 at 0:46
  • $\begingroup$ 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 $\endgroup$ – Sum'math Feb 18 '17 at 1:05
  • $\begingroup$ Using that, you can then derive the closed form of your sum as martycohen has laid out for you. $\endgroup$ – Simply Beautiful Art Feb 18 '17 at 1:06
  • $\begingroup$ Ah! I have worked it out now on my own $\endgroup$ – Sum'math Feb 18 '17 at 1:31
  • $\begingroup$ Wonderful! And welcome to the site :D $\endgroup$ – Simply Beautiful Art Feb 18 '17 at 1:35
3
$\begingroup$

$\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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.