1
$\begingroup$

On Wikipedia the following closed form is derived - Generalised formula

enter image description here

Can someone explain how the closed form below is derived?

enter image description here

Edit

Solution thanks to graydad

enter image description here

$\endgroup$

1 Answer 1

2
$\begingroup$

The first one is a pretty standard exercise. Let $S_k = \sum_{k=a}^b r^k$. $\space$Then $$rS_k = r\sum_{k=a}^b r^k = \sum_{k=a}^b r^{k+1}$$ so $$rS_k-S_k = (r^{a+1}+r^{a+2}+ \dots + r^{b+1})- (r^{a}+r^{a+1}+ \dots + r^{b}) \\ = r^{b+1}-r^{a}$$ and since $$rS_k-S_k = S_k(r-1)$$ we have enough to know that $$S_k(r-1)=r^{b+1}-r^{a} \\ \implies S_k = \frac{r^{b+1}-r^{a}}{r-1}$$ Try to use a similar tactic in deriving the second equality you are interested in, or just plug in $r = \frac{1}{\hat{r}}$ and solve in terms of $\hat{r}$ so that it looks like the second equation.

$\endgroup$
5
  • $\begingroup$ Thanks, not sure how to turn k negative. $\endgroup$ Nov 26, 2014 at 16:18
  • $\begingroup$ $r^{-k} = \frac{1}{r^k}$. Try making the substitution I mention in my last line. $\endgroup$
    – graydad
    Nov 26, 2014 at 16:19
  • $\begingroup$ Did you get it yet? $\endgroup$
    – graydad
    Nov 26, 2014 at 17:46
  • $\begingroup$ Yes, thanks. :-) $\endgroup$ Nov 26, 2014 at 20:12
  • $\begingroup$ Great! Please accept my answer so that other users know this question is closed :) Or take the solution you added above and accept it as your own answer to this question. $\endgroup$
    – graydad
    Nov 26, 2014 at 20:16

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.