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October 8th, 2016, 05:21 PM  #1 
Newbie Joined: Jan 2012 Posts: 2 Thanks: 0  Reverse an equation containing a sum
Hello, I currently have the following equation: $\displaystyle y = (\sum_{n=1}^{z} (n + 300 * 2^{{z}/{7}})) / 4$ If we give us z, I have no trouble finding y, but if we get y how can we get z? Having trouble reversing the equation... Trying to isolate z, but no idea on how to proceed with the sum. $\displaystyle 4 * y = \sum_{n=1}^{z} (n + 300 * 2^{z/7})$ 
October 8th, 2016, 06:39 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
I don't think the final term is correct. $$\sum_{n=1}^{7m} 2^m = 2\sum_{n=0}^{7m1} 2^m$$ $$\begin{align*} \text{if} \quad S &= 2(1+2+4+8+\ldots+2^{7m1}) \\ 2S &= 2(\phantom{1+}2+4+8+\ldots+2^{7m1}+2^{7m}) \\ S &= 2(2^{7m} 1) \\ \end{align*}$$ Thus, with $z=7m$ we have the final terms as $75 \times 2(2^z1)$. 
October 8th, 2016, 06:46 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 2,500 Thanks: 1372 
thought I had deleted it

October 8th, 2016, 08:44 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
I must have got there before you did. Looking at it, what I wrote isn't correct either. Your one probably was (although I can't remember it exactly). 
October 8th, 2016, 08:48 PM  #5 
Senior Member Joined: Sep 2015 From: USA Posts: 2,500 Thanks: 1372  It agreed with Mathematica, but when I saw that there was no closed form solution I meant to leave it for someone else to take a look at and thought I deleted it.
Last edited by skipjack; October 9th, 2016 at 06:02 AM. 

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