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I am having trouble with this equation in a class I am taking and I am trying to understand it: $$\sum_{i=47}^{i=136} M_i$$

We have to solve for the problem below but the hint our professor gave us was subtraction. I am confused because it is my understanding that the bottom was the starting point and the top was the ending point. So wouldn't $M_i$ start with $M_{47}$ and work up from there? Any help is much appreciated.

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    $\begingroup$ How is $M_i$ defined? We don't have enough information to answer otherwise. $\endgroup$ – jwc845 Sep 25 '18 at 16:02
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The summation can be rewritten as:

$$\sum_{i=47}^{i=136} M_i =\sum_{i=0}^{i=136} M_i - \sum_{i=0}^{i=46} M_i$$

The reason behind this is that the when summing from $i=0$ to $i=136$ we would be also including all $M_i$ from $i=0$ to $i=46$. We can then use subtract the sum of all numbers in that range to get the final result.

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  • $\begingroup$ Hey, thank you for the explanation there! I really do appreciate it. So he is stating to find that summation. Would that rewritten summation be the answer at that point? I apologize I am struggling a little to understand this stuff lol. $\endgroup$ – Schmit Sep 25 '18 at 16:26
  • $\begingroup$ He probably wants you to calculate the value of the rewritten summation which I can't do without knowing what $M_i$ is defined as. The rewriting helps because depending on what your $M_i$ means it is probably easier to calculate the sum from 0 to 136 and 0 to 46 and subtracting than the alternative. Also the bottom of the summation may make more sense as $i=1$. $\endgroup$ – jwc845 Sep 25 '18 at 17:05
  • $\begingroup$ Well, that is the issue, he does not define or give a value for M. Which is part of why I was confused in the first place haha. $\endgroup$ – Schmit Sep 25 '18 at 17:17
  • $\begingroup$ Maybe he means $M_i=I$, so it be (47+48+49...136)? This would be where the simplification of 2 sums starting at 0 would help because there is a known formula for the sum of the first $n$ integers. $\endgroup$ – jwc845 Sep 25 '18 at 17:39
  • $\begingroup$ Yeah, I did not think of that. So if that was the case, I would have (47, 48, 49...136) and then (1, 2, 3, 4...136) and then use the rewritten summation formula written above? $\endgroup$ – Schmit Sep 25 '18 at 18:06
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You are right, the summation you are given is over the sequence 47, 48, 49, $\dots$, 136.

You are probably used to summation formulas that start with 1. So can you rewrite the given summation in terms of those familiar ones? $$ \sum_{i=47}^{136} M_i = \sum_{i=1}^{?} M_i - \sum_{i=1}^{?} M_i $$

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Without knowing what $M_i$ is, I can only guess that he might be suggesting that you do $$\sum_{i=47}^{i=136} M_i = \sum _{i=0}^{i=136} M_i - \sum_{i=0}^{i=46}M_i $$

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