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 :)

  • $\begingroup$ 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

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

Then use sum of GP.


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