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How to establish the formula for this without induction?

$$1 + 2 + 3 + · · · + n =\sum_{j=1}^n j=\frac{n(n + 1)}{2}$$

Peter
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maliiaButterfly
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    In the case $n$ is even, use the Gauss-trick : Pair $1$ with $n$, $2$ with $n-1$ and so on. All $\frac{n}{2}$ pairs have sum $n+1$. In the odd case, things are slightly more complicated. – Peter Aug 03 '17 at 18:25

7 Answers7

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$1+2+.....+n$

$n+n-1+...+1$

add the both lines you hobtain $2S_n=n(n+1)$ and $S_n=n(n+1)/2$.

Tsemo Aristide
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Hint :

Define $$S = 1 + 2 + \dots + n.$$ We can write $$2S = 1 + 2 + \dots + n + n + \dots + 2 + 1$$ So what is $S$ now?

Peter
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Let $$S=1+2+3+\cdots+n$$ Reversing the terms and adding it to itself produces $$\begin{align*}2S &=\overbrace{(n+1)+(n+1)+\cdots+(n+1)}^n\\&=n(n+1)\end{align*}$$ Hence $$S=\frac{n(n+1)}{2}$$

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Proof without words (from the Stanford Encyclopedia of Philosophy):

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Alex Ortiz
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There are actually quite a few ways of writing this up. I will not detail them all out here, but I covered quite a few ways in this posting of mine for my students:

https://cgmcwhor.expressions.syr.edu/wp-content/uploads/2015/01/on_the_sum_1-n.pdf

I present a few different ways:

$1$. By induction.

$2$. By 'dots'

$3$. Via Gauss

$4$. Using summations

$5$. Using combinatorics

$6$. Using binomial coefficients

$7$. Using Graph Theory

$8$. Using Triangles

$9$. Using l'Hopitals

$10$. Using Telescoping Series

mathematics2x2life
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  • Hi, I don't understand why does x has to approach 1 while taking the limit in l'Hopital method? Why does it not approach any other number, why 1? – maliiaButterfly Aug 04 '17 at 03:56
  • @maliiaButterfly Notice we differentiated the sum $1+r+\cdots+r^n$ to obtain $1+2r+3r^2+\cdots+nr^{n-1}$. So if $r=1$, then we obtain the sum $1+2+3+\cdots+n$. However, we must take the limit as $r \to 1$ since the equality isn't valid for $r=1$ so we must be careful about the behavior at $r=1$. But notice, the limit exists and works out to be just what we want, namely $\dfrac{n(n+1)}{2}$. – mathematics2x2life Aug 04 '17 at 04:36
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The arithmetic mean of the first $n$ natural numbers is clearly $(n+1)/2$. Since there are $n$ of them, the sum is $n(n+1)/2$.

ajotatxe
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First proof: $$ \begin{align} S&=1+2+3+...+n \\ S&=n+(n-1)+(n-2)+...+1\\ 2S&=(n+1)+(n+1)+(n+1)+...(n+1)=n(n+1) \end{align} $$

My favourite proof(Combinatorial):

Suppose there are $n+1$ person in a gathering and every one will shake hand with everyone else exactly once. What is the total no of handshake?

I will count it in two different ways.

First way: Every person shakes hand $n$ times. So there may be $n(n+1)$ handshakes. But we have counted every handshake exactly twice. So the total no of handshake is $\frac{n(n+1)}{2}$.

Second way: Put all $n+1$ persons in a row and number them from $1$ to $n+1$. To avoid double counting each person shakes hand with persons with higher number ie. person $1$ shakes hand with 2 to n+1 and leaves, then person 2 shakes hand with 3 to n+1 and leaves and so on. So the total number of handshakes is $$\sum_{i=1}^{n+1} (n+1)-i=\sum_{i=0}^n (n-i)$$

J. Sarkar
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