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Possible Duplicate:
Sum of n consecutive numbers

I really can't remember (if I have ever known this): which series is this and how to demonstrate its solution?

$\sum\limits_{i=1}^n i = \frac{n(n+1)}{2}$

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    @$B$rad You are right, the very definition of series is a bit ambiguous. I was just joking.2012-08-10

4 Answers 4

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$S = 1 + 2 + 3 + \ldots + n = n + (n-1) + (n-2) + \ldots + 1$. So $2S = (n+1) + \ldots + (n+1)$. Since the $(n+1)$ appears $n$ times, $S = \frac{n(n+1)}{2}$

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Write out this sum twice, once is direct order, and once in reverse: $ \begin{align} &1 + &2 + &\ldots+ &(n-1)+ &n &=s \\ &n + &(n-1)+ & \ldots+ &2+ &1 &=s \end{align} $ Now add up column-wise: $ (n+1) + (n+1) + \ldots + (n+1) + (n+1) = 2s $ There are exactly $n$ terms here (as many as the number of terms in the sum). Hence: $ n(n+1) = 2s $ Now solve for $s$.

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This is an Arithmetic Series starting from $1$ with difference $1$. $ \sum\limits_{i=1}^n i = 1 + 2 + 3 + 4 + ... +n = {n (n+1) \over 2}$ 1#Check this out.
2#There's one another related one.

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This is not a series. This sum is named Gauss sum and that formula $\frac{n(n+1)}{2}$ you can prove it using induction.

The exercise starts from the following sum: $1+2+ \ldots +100$ and the way you can classify the terms of this sum.

$1+2+ \ldots + 100 = (1+100)+ (2+99)+ \ldots (50+51$).

for more information I think the following link : http://www.newton.dep.anl.gov/askasci/math99/math99155.htm it is a good one.

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    As @Brad pointed out, the term "series" is not completely wrong. A series is the sequence obtained by summing the terms of a given sequence. In this case, the given sequence is $\{1,2,3,4,\ldots,n,0,0,\ldots\}$.2012-08-10