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I have two examples of sigma summation, where $n$ & $j$ are positive integers and $c$ & $x$ are any real or complex numbers.

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The first makes complete sense to me, however; the second partially doesn't. So, for i) I understand that I am just adding the same number to itself $n$ times hence $nc$, applying the same to ii) $x^j$ is just another constant and adding itself over and over up until I reach $n$ implies to me that it's equal to $nx^j$. My only reasoning as to why this wouldn't be true is because I am not negating the "zeroth" term hence the need to use $n+1$. Am I correct on my later assumption?

Lo-urc
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Yes, you are broadly correct. Your summations are adding up the same number, some number of times. So, the total will be $$(\# \text{ times you are adding it up})\times(\text{value you are adding}).$$

In (i) you have a summand for each number between $k=1$ and $n$ inclusive, which is $n$ many, and the value is $c$: hence $nc$. In (ii) you have a summand for each number between $k=0$ and $n$, which is $n+1$ many, and the value is $x^j$: hence $(n+1)x^j$.

James
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  • I have also come across one more Im having trouble understanding. I understand the first part and the middle, however the final form that it is expressed in I don't understand how could be achieved. https://goo.gl/pnD7yK – Lo-urc Oct 13 '17 at 20:37
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    @Jan Remember that $\sum_{i=0}^n x_i = x_0 + x_1 + x_2 + \cdots + x_{n-1} + x_n$ and $\sum_{i=1}^n x_i = x_1 + x_2 + \cdots + x_{n-1} + x_n$ and $\sum_{i=0}^{n-1} x_i = x_0 + x_1 + x_2 + \cdots + x_{n-1}$. – James Oct 13 '17 at 20:43
  • I have this issue of reading things differently despite it also being correct and it's really annoying, is this summation read as the left or right form? https://goo.gl/7gY3PM (Left makes sense) – Lo-urc Oct 13 '17 at 20:52