0
$\begingroup$

Earlier I was looking to find a closed form expression of the sum: $$\sum_{h=1}^{k-1}\left({2^{h}}{3^{k-h-1}}\right)$$ Wolfram Alpha tells me that this is equivalent to: $$2\times{3^{k-1}}-2^k$$ However it does not explain the method of how it got there, and I cannot seem to find any resources online explaining how expressions can be found for more general cases (although I'm sure they're out there). I was wondering if anyone could give a method for deriving this or point to resources that explain how different sums can be expressed in closed form.

Thanks :)

$\endgroup$
  • $\begingroup$ It's part of an expression I found for the $n$th iteration of $\frac{3n+1}{2}$, so now I know that it's $({\frac{3}{2}})^k\left(n+1\right)-1$, hence the number of iterations of the Collatz function for odd $n$ to become even is one less than the power of two in the prime factorization of $n+1$ $\endgroup$ – Daniel Castle Jul 3 '17 at 1:19
  • $\begingroup$ I doubt it, I just wanted to investigate the problem a bit after I saw it on numberphile $\endgroup$ – Daniel Castle Jul 3 '17 at 1:21
  • $\begingroup$ But the formula works, I've tested it. It must have already be found, surely... $\endgroup$ – Daniel Castle Jul 3 '17 at 1:22
  • $\begingroup$ I also found that all numbers of the form ${(\frac{2}{3})}^{a}\cdot(2^{3^{a-1}\cdot{(2b-1)}}+1)-1$ where a and b are positive integers definitely becomes a power of 2 after a iterations, if it's of any use. $\endgroup$ – Daniel Castle Jul 3 '17 at 1:29
  • $\begingroup$ Just did, I suck at MathJax.. $\endgroup$ – Daniel Castle Jul 3 '17 at 1:30
3
$\begingroup$

$$2^h3^{k-h-1} = 3^{k-1}\left(\frac{2}{3}\right)^h$$

Then use sum of GP.

$\endgroup$

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.