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This question already has an answer here:

Find the sum $\sum_{j=0}^{n}j$

for all $n=0,1,2,3,\dots$.

How do I find/evaluate this sum?

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marked as duplicate by Pedro Tamaroff real-analysis Nov 20 '15 at 23:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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HINT

  • add first and last term to get $n+1$
  • next pair (next to last and second) gives the same result
  • see Gauss Trick
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  • $\begingroup$ I don't get your first hint, could you elaborate? $\endgroup$ – jukka.aalto Nov 4 '15 at 14:59
  • $\begingroup$ @jukka.aalto first term in your sum and last term added together are $n+1$. Next pair is $2 + (n-1) = n+1$. etc $\endgroup$ – gt6989b Nov 4 '15 at 15:27
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Consider: $$S = \sum_{j=0}^n x^j = \frac{x^{n+1} - 1}{x - 1}$$ $$\frac {d}{dx}S = \sum_{j=0}^njx^{j-1} = \frac{nx^{n+1} - (n+1)x^n + 1}{(x-1)^2}$$

If we take the limit as $x \to 1$, the left hand side becomes $\sum_{j=0}^n j$, and we can evaluate the RHS to find a function of $n$ $$\lim_{x\to 1}\frac{nx^{n+1} - (n+1)x^n + 1}{(x-1)^2} = \lim_{x\to1}\frac{n(n+1)x^n - (n+1)nx^{n-1}}{2(x-1)}$$ $$= \lim_{x\to1}\frac{n(n+1)x^n - (n+1)nx^{n-1}}{2(x-1)} = \lim_{x\to1}x^{n-1}\frac{(n+1)n}{2} = \frac{(n+1)n}{2}$$

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  • $\begingroup$ This is an obscure way to derive a basic result. $\endgroup$ – Simon S Nov 4 '15 at 15:15
  • $\begingroup$ @SimonS It was more of a sarcastic answer. In fact I'm surprised no one has flagged this question as a duplicate yet $\endgroup$ – jameselmore Nov 4 '15 at 15:24
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Here's a proof without words that should help, just generalize from there.

enter image description here

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Here is a high school argument which I love more! :)

Assume $n$ is an odd positive integer then

$$\eqalign{ & \sum\limits_{j = 0}^n j = 0 + 1 + 2 + ... + (n - 2) + (n - 1) + n \cr & \,\,\,\,\,\,\,\,\,\,\, = \left( {0 + n} \right) + \left( {1 + \left( {n - 1} \right)} \right) + \left( {2 + \left( {n - 2} \right)} \right) + ... \cr & \,\,\,\,\,\,\,\,\,\,\, = \underbrace {n + n + n + ...}_{{{n + 1} \over 2}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \left( {{{n + 1} \over 2}} \right)n = {{n(n + 1)} \over 2} \cr} $$

you can argue most similarly when $n$ is an even number! I will leave this case as an exercise for you. :)

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