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Proof for formula for sum of sequence $1+2+3+\ldots+n$?

How was this derived as the sum of the following sequence...

$$S_n = (n–1)+(n–2)+(n–3)+\dotsb+1=\frac{n\cdot(n–1)}{2}$$

Please could you guide me in the right direction...

$$\begin{aligned} S_n &= 1+\dotsb+ (n-3)+(n-2)+ (n-1) \\ S_n &= (n–1)+(n–2)+(n–3)+\dotsb+1 \\ 2S_n &= (n-1+1)+(n-2+n-3)+(n-3+n-2)+(n-1+1) \\ S_n &= \dotsc \end{aligned}$$

I'm not sure where to go from here...

Edit solved: forgot to times n by last term.

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Read here, link,- nice story about young Gauss. –  Salech Alhasov Mar 3 '12 at 19:53
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marked as duplicate by Ilmari Karonen, Eric Naslund, Kannappan Sampath, mixedmath, Henry Mar 3 '12 at 20:39

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3 Answers

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Write $S_n$ in two ways, add them up... $$\begin{array}{cccccc} S_n &=& 1 &+& 2 &+& \ldots &+& (n-2) &+& (n-1) \\ S_n &=& (n-1) &+& (n-2) &+& \ldots &+& 2 &+& 1 \\ \hline 2S_n &=& 1 + (n - 1) &+& 2 + (n - 2) &+& \ldots &+& (n - 2) + 2 &+& (n - 1) + 1 \\ &=& n &+& n &+& \ldots &+& n &+& n \\ &=& n (n - 1) & & & & & & & & \end{array}$$ ...and divide both sides by $2$ to get $S_n = n(n-1)/2$.

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You almost had it, but you have things lined up wrong.

$S_n=1+2+3+...+(n-3)+(n-2)+(n-1)$

$S_n=(n-1)+(n-2)+(n-3)+...+3+2+1$

$2S_n=(n-1+1)+(n-2+2)+...$

Maybe someone else can line that up better, but you should see where this is going.

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@Hmm Where's that extra 1 coming from? –  Mike Mar 3 '12 at 20:10
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Hint: Think about triangular numbers. Specifically, consider the difference $S_{n+1} - S_n$.

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Really? Does that have the same difference between successive terms as you original sequence? –  Ilmari Karonen Mar 3 '12 at 19:59
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