# Doubt about series - which series is this? [duplicate]

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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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## marked as duplicate by Andrés E. Caicedo, Sasha, David Mitra, copper.hat, Asaf KaragilaAug 10 '12 at 15:43

This is not a series, since there is no infinite sum :-) It is a summation identity, rather standard, and can be easily proved by mathematical induction on $n$. This is indeed one of the most popular exercises on mathematical induction. – Siminore Aug 10 '12 at 15:17
@Siminore: The term "series" is commonly used for both finite and infinite summations. – Brad Aug 10 '12 at 15:21
By the way, there is a beautiful visual proof of this fact appearing as the top answer to this MO question: mathoverflow.net/questions/8846/proofs-without-words – Brad Aug 10 '12 at 15:22
@Brad You are right, the very definition of series is a bit ambiguous. I was just joking. – Siminore Aug 10 '12 at 15:29

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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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 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.

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

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 $\{1,2,3,4,\ldots,n,0,0,\ldots\}$. – Siminore Aug 10 '12 at 15:29
$S = 1 + 2 + 3 + \ldots + n = n + (n-1) + (n-2) + \ldots + 1$. So $2S = (n+1) + \ldots + (n+1)$. Since the $(n+1)$ appears $n$ times, $$S = \frac{n(n+1)}{2}$$