Sum of the first natural numbers: how many and what are the most common methods to verify it? We know that Gauss has shown that the sum $S$ of the first $n$ natural numbers is given by the relation:
$$S=\frac{n(n+1)}{2} \tag{*}$$
The proof that I remember most frequently is as follows:
Let be $S=1+2+\dotsb+(n-1)+n \tag{1}$ We can write $S$ it also as: $\tag{2} S=n+(n-1)+\dotsb+2+1.$
By adding up member to member we get:
$\tag{3} 2S=\underbrace{(n+1)+(n+1)+\dotsb+2+1}_{n-\mathrm{times}}.$
Hence we obtain the $(^\ast)$. 
How many other simple methods exist to calculate the sum of the first natural numbers?
 A: Here is a proof using ideas from probability theory. Let $X$ be a random variable that is uniformly distributed on the set $\{1,\dotsc, n\}$. Then it is easy to see that $n+1-X$ has the same distribution as $X$ whence
$$
E(n+1-X)=EX
$$
i.e. $n+1-EX=EX$ so 
$$
EX=\frac{n+1}{2}\tag{0}.
$$
But on the other hand,
$$
EX=\sum_{k=1}^n kP(X=k)=\frac{1}{n}\sum_{k=1}^n k.\tag{1}
$$
Combining $(0)$ and $(1)$ we get that
$$
\sum_{k=1}^nk=\frac{n(n+1)}{2}.
$$
A: The identity
$$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$$
is a classical result which can be easily proved by the following 

A: The expression of $S_n$ must be a quadratic polynomial (because the difference $S_n-S_{n-1}=n$ is a linear polynomial). As $S_0=0$, it is of the form $an^2+bn.$
By identification,
$$\begin{cases}S_1=a+b=1,\\S_2=4a+2b=3\end{cases}$$
and $$a=b=\dfrac12.$$

The Lagrangian polynomial by $(0,0)$, $(1,1)$, $(2,3)$ is $\dfrac{x^2}2+\dfrac x2$.

$$S(n)-S(n-1)=an^2+bn-a(n-1)^2-b(n-1)=a(2n-1)+b=n$$
so that
$$a=b=\dfrac12.$$

A posteriori:
$$S_n-S_{n-1}=\frac{n(n+1)}2-\frac{(n-1)n}2=n.$$

As an arithmetic progression is linear, the average term is the average of the extreme terms.
$$\frac{S_n}n=\frac{1+n}2$$ and $$S_n=\frac{n(n+1)}2.$$

Let the function defined by a geometric series
$$f(x):=\sum_{k=0}^n x^k=\frac{x^{n+1}-1}{x-1}.$$
We differentiate on $x$,
$$f'(x)=\sum_{k=0}^n k x^{k-1}=\frac{x^{n+1}-1}{x-1}=\frac{(n+1)x^n(x-1)-(x^{n+1}-1)}{(x-1)^2}.$$
Then with $x=1$, using L'Hospital,
$$f'(1)=\sum_{k=0}^n k =\lim_{x\to0}\frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2} =\lim_{x\to0}\frac{n(n+1)x^{n}-(n+1)nx^{n-1}}{2(x-1)}=\frac{n(n+1)}2.$$
A: Certainly overkill, but we can do this with the method of generating functions: let
$$f(x)=\sum_{k=0}^nx^n=\frac{1-x^{n+1}}{1-x}$$
So that $f'(1)=\sum_{k=0}^nk$. We just need to find $f'(x)$, and we proceed as usual:
$$\frac{d}{dx}\frac{1-x^{n+1}}{1-x}=\frac{-(n+1)x^n(1-x)+(1-x^{n+1})}{(1-x)^2}$$
Taking the limit as $x\to1$, we have by L'hopital's rule and after simplification,
$$\lim_{x\to1}\frac{-n(n+1)x^{n-1}(1-x)}{-2(1-x)}=\lim_{x\to1}\frac{n(n+1)}{2}=\frac{n(n+1)}{2}$$
From which we obtain the conclusion.
A: Use can use the sum of AP to arrive at the same conclusion.
$a=1, d=1$
$$S_n = {n\over 2}(2a+(n-1)d)$$
$$S_n = {n\over 2}(2+(n-1))$$
$$S_n = {n\over 2}(n+1)$$
