0
$\begingroup$

Possible Duplicate:
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.

$\endgroup$

marked as duplicate by Ilmari Karonen, Eric Naslund, user21436, davidlowryduda, Henry Mar 3 '12 at 20:39

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.

  • $\begingroup$ Read here, link,- nice story about young Gauss. $\endgroup$ – Salech Rubenstein Mar 3 '12 at 19:53
  • $\begingroup$ Proof by induction is also possible. $\endgroup$ – TurlocTheRed Sep 25 '18 at 18:05
0
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @Hmm Where's that extra 1 coming from? $\endgroup$ – Mike Mar 3 '12 at 20:10
0
$\begingroup$

Hint: Think about triangular numbers. Specifically, consider the difference $S_{n+1} - S_n$.

$\endgroup$

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