# How to Solve Summation Notation

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.

• How is $M_i$ defined? We don't have enough information to answer otherwise. – jwc845 Sep 25 '18 at 16:02

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.

• 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. – Schmit Sep 25 '18 at 16:26
• 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 $i=1$. – jwc845 Sep 25 '18 at 17:05
• 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. – Schmit Sep 25 '18 at 17:17
• 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. – jwc845 Sep 25 '18 at 17:39
• 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? – Schmit Sep 25 '18 at 18:06

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? $$\sum_{i=47}^{136} M_i = \sum_{i=1}^{?} M_i - \sum_{i=1}^{?} M_i$$

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