Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

3 Answers

up vote 4 down vote accepted

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

share|improve this answer
    
Didn't expect answers which would come as quickly and be as good. Thanks. –  TheCoconutChef Jul 7 '12 at 3:13
add comment

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

share|improve this answer
    
Thanks. Quite helpful. –  TheCoconutChef Jul 7 '12 at 3:13
add comment

\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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.