My Math Forum Reverse an equation containing a sum

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 October 8th, 2016, 06:21 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, 07:39 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,700 Thanks: 2682 Math Focus: Mainly analysis and algebra I don't think the final term is correct. $$\sum_{n=1}^{7m} 2^m = 2\sum_{n=0}^{7m-1} 2^m$$ \begin{align*} \text{if} \quad S &= 2(1+2+4+8+\ldots+2^{7m-1}) \\ 2S &= 2(\phantom{1+}2+4+8+\ldots+2^{7m-1}+2^{7m}) \\ S &= 2(2^{7m} -1) \\ \end{align*} Thus, with $z=7m$ we have the final terms as $75 \times 2(2^z-1)$.
 October 8th, 2016, 07:46 PM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,647 Thanks: 1476 thought I had deleted it
 October 8th, 2016, 09:44 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,700 Thanks: 2682 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, 09:48 PM   #5
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Joined: Sep 2015
From: USA

Posts: 2,647
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Quote:
 Originally Posted by v8archie 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).
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 07:02 AM.

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