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 :)

  • $\begingroup$ what makes you assume that it has a closed form? Also, what have you tried. Also, why do you want an answer? Why should anyone care? You have provided no justification on why this question deserves answering. $\endgroup$ – The Great Duck Jul 3 '17 at 1:12
  • $\begingroup$ 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
  • $\begingroup$ Wait... this actually simplifies the collatz conjecture? $\endgroup$ – The Great Duck Jul 3 '17 at 1:20
  • $\begingroup$ 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
  • $\begingroup$ actually, finding a formula for the nth iteration of (3n+1)/2 would be useful. $\endgroup$ – The Great Duck Jul 3 '17 at 1:22

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

Then use sum of GP.


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