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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?


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

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

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


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


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.

  • $\begingroup$ @ 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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