Without using the proof by induction establish the formula for the simple arithmetic series How to establish the formula for this without induction?
$$1 + 2 + 3 + · · · + n =\sum_{j=1}^n j=\frac{n(n + 1)}{2}$$
 A: $1+2+.....+n$
$n+n-1+...+1$
add the both lines you hobtain $2S_n=n(n+1)$ and $S_n=n(n+1)/2$.
A: Hint : 
Define $$S = 1 + 2 + \dots + n.$$ We can write $$2S = 1 + 2 + \dots + n + n + \dots + 2 + 1$$ So what is $S$ now?
A: 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 $$S=\frac{n(n+1)}{2}$$
A: Proof without words (from the Stanford Encyclopedia of Philosophy):

A: 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
A: The arithmetic mean of the first $n$ natural numbers is clearly $(n+1)/2$. Since there are $n$ of them, the sum is $n(n+1)/2$.
A: 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 $$\sum_{i=1}^{n+1} (n+1)-i=\sum_{i=0}^n (n-i)$$
