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This question already has an answer here:

Could somebody explain me how to sum the following series

$$ (x-1) + (x-2) + ... + 1 = ? $$

I got above series as a part of equation and was wondering how to simplify it. As a matter of fact, we have arithmetic progression here which can be easily summarized.

I used wolframalpha to calculate the sum like this.

However, wolfram did something which I don't understand. It simplified the equation and gave me the following result:

$$ \sum_{k=0}^{x} \big(1-k \big) = -\frac{1}{2} (x-2)(x+1) $$

Don't see how to get the right-hand side of the sum. It looks like there is a theory behind the scene, but I don't know where to start looking from. My guess is it somehow connected with generating functions, but I'm not sure.

Can somebody hint me?

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marked as duplicate by Alex M., Lucian, S.C.B., kingW3, Henrik Jan 12 '17 at 18:13

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.

  • $\begingroup$ I think this question might be closed due to duplicate, so I just give a comment: Search for arithmetic sum. $\endgroup$ – mickep Jan 12 '17 at 7:44
  • $\begingroup$ The first sum you show us and the second one you wrote are very different! Was that on purpose? $\endgroup$ – RGS Jan 12 '17 at 7:45
  • $\begingroup$ @RSerrao no, I've just fixed it. $\endgroup$ – Roman Dryndik Jan 12 '17 at 7:47
  • $\begingroup$ I just checkedthe image. WolframAlpha did not understand what you mean amd gave you the wrong answer. $\endgroup$ – RGS Jan 12 '17 at 7:47
  • $\begingroup$ Your post is contradictory. You say that you know how to handle an arithmetic progression, and then that you don't understand WA's result, that is exactly applying the formula. $\endgroup$ – Yves Daoust Jan 12 '17 at 8:02
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Looking at the series again after rewriting the last term, we get$$(x-1)+(x-2)+...+(x-(x-1))$$

WolframAlpha's solution is wrong as it failed to interpret the series.

It should be$$\sum_{k = 1}^{x-1}x-k = \frac {x(x-1)}{2}$$

P.S.: If you don't know how to get the R.H.S., see below.

Its an A.P. with first term $(x-1)$, common difference $-1$ and last term $1$.

What we need to find out is the number of terms here. Looking at the series again,$$(x-1)+(x-2)+...+(x-(x-1))$$ As we can see, number of terms $$n = x-1$$

Now, we know that sum of an A.P. is$$S = \frac{n}{2}(a+l)$$where $l$ is the last term and other terms have their usual meanings $$S=\frac{x-1}{2}(x-1+1)$$

$$S=\frac {x(x-1)}{2}$$

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Hint: If $a_n$ is an arithmetic progression,and $\displaystyle S_n=\sum_{k=1}^{n}a_k$, then we have $$S_n=\frac{\left ( a_1+a_n \right )n}{2}$$

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Here's a proof that sort of uses generating function techniques. Consider the polynomial $p(x) = 1 + x + x^2 + ... + x^n = \frac{x^n-1}{x-1}$. Take the derivative $p'(x) = 1 + 2x + 3x^2 +... +nx^{n-1} = \frac{(n-1)x^n -nx^{n-1} + 1}{(x-1)^2}$. You can use L'Hopitals rule to take the limit as $x \to 1$ of the right hand side of the equality, thus finding $p'(1) = 1 + 2 +...+n = \frac{n(n+1)}{2}$, which is equivalent to the formula given by Wolfram. Note that this is a very widely known result which is usually proven in a more elementary way.

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  • $\begingroup$ @ Vik78 I see arithmetic progression here. But its sum is $ \frac{(x-1+1)}{2} \cdot (x-1) = \frac{x(x-1)}{2} $. Sorry, in my previous post I've made wrong latex markup, so the formula weren't properly rendered. $\endgroup$ – Roman Dryndik Jan 12 '17 at 8:07
  • $\begingroup$ Just substitute $x -1 = n$ into my formula to see that this is exactly the answer I gave. $\endgroup$ – Vik78 Jan 12 '17 at 8:10
  • $\begingroup$ I've tried it, but maybe I'm doing something wrong. Assuming n = x - 1, we have another representation of the series. And $ p'(1) = 1 + 2 +...+n = \frac{n(n+1)}{2} = \frac{(x-1) \cdot x}{2} $ $\endgroup$ – Roman Dryndik Jan 12 '17 at 8:21
  • $\begingroup$ Yes, so $1 + 2 +... + n = 1 + 2 +... + x-1 = \frac{(x-1)x}{2}$. What's the issue? This seems to be exactly the answer you expected. $\endgroup$ – Vik78 Jan 12 '17 at 8:25
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    $\begingroup$ Dude, this is literally the exact answer I gave you like five times now. $\endgroup$ – Vik78 Jan 12 '17 at 9:07

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