# Understanding this summation identity

I'm currently reading a book in which part of the solution to the problem involve this identity:

$$\sum_{j=i+1}^{n}j = \sum_{j=1}^{n}j-\sum_{j=1}^{i}j$$

Which I cannot derive myself. The only thing I can do with it is this:

$$\sum_{j=i+1}^{n}j = \sum_{j=1}^{n}j+i = \sum_{j=1}^{n}j + \sum_{j=1}^{i}i$$

Which seems to me completely useless.

Any help in understanding this (as I am unaccustomed to summation manipulation in general) would be greatly appreciated.

I know it's related to " Calculate integer summation when lower bound is a variable " but I still don't see the why.

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We have $$\sum_{j=1}^{i}j+\sum_{j=i+1}^n j=\sum_{j=1}^n j.\tag{1}$$ The result you are looking for follows by subtraction.

If $(1)$ seems unclear, let us take particular numbers, say $i=7$ and $n=19$. We have $$\sum_{j=1}^{i}j =\sum_{j=1}^7j=1+2+\cdots+7$$ and $$\sum_{j=i+1}^{n}j =\sum_{j=8}^{19} j=8+9+\cdots +19.$$ If you add them, you get $$1+2+\cdots+7+8+9+\cdots +19.$$ This is equal to $$\sum_{j=1}^{19} j.$$

Remark: Exactly the same argument shows that if $a_1,a_2,\dots$ is any sequence, then $$\sum_{j=i+1}^n a_j=\sum_{j=1}^n a_j -\sum_{j=1}^i a_j.$$

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Didn't expect answers which would come as quickly and be as good. Thanks. – TheCoconutChef Jul 7 '12 at 3:13

Note that \begin{align} \sum_{j=1}^{n} j & = 1 + 2 + \cdots + i + (i+1) + (i+2) + \cdots + n\\ & = \left(1 + 2 + 3 + \cdots + i \right) + \left((i+1) + (i+2) + \cdots + n \right)\\ & = \sum_{j=1}^{i} j + \sum_{j=i+1}^{n} j \end{align} Hence, we get that $$\sum_{j=i+1}^{n} j = \sum_{j=1}^{n} j - \sum_{j=1}^{i} j$$

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Thanks. Quite helpful. – TheCoconutChef Jul 7 '12 at 3:13

\begin{align} 6+7+8 = \Big(1+2+3+4+5+6+7+8\Big) - \Big(1+2+3+4+5\Big) \end{align}

That's all there is to it. Your way of phrasing the question makes me wonder if you think there's something more to it than that.

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I guess that when manipulating new symbols (the Sigma notation isn't exactly new to me but it had never been center stage) their complete meaning isn't immediately apparent. I didn't see it concretely. – TheCoconutChef Jul 7 '12 at 12:51