Bsrgvty

We know that Gauss has shown that the sum $S$ of the first $n$ natural numbers is given by the relation:

$$S=fracn(n+1)2 tag*$$
The proof that I remember most frequently is as follows:

Let be $S=1+2+dotsb+(n-1)+n tag1$ We can write $S$ it also as: $tag2 S=n+(n-1)+dotsb+2+1.$
By adding up member to member we get:
$tag3 2S=underbrace(n+1)+(n+1)+dotsb+2+1_n-mathrmtimes.$
Hence we obtain the $(^ast)$.

How many other simple methods exist to calculate the sum of the first natural numbers?

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
$$
EX=fracn+12tag0.
$$
But on the other hand,
$$
EX=sum_k=1^n kP(X=k)=frac1nsum_k=1^n k.tag1
$$
Combining $(0)$ and $(1)$ we get that
$$
sum_k=1^nk=fracn(n+1)2.
$$

Certainly overkill, but we can do this with the method of generating functions: let
$$f(x)=sum_k=0^nx^n=frac1-x^n+11-x$$
So that $f'(1)=sum_k=0^nk$. We just need to find $f'(x)$, and we proceed as usual:
$$fracddxfrac1-x^n+11-x=frac-(n+1)x^n(1-x)+(1-x^n+1)(1-x)^2$$
Taking the limit as $xto1$, we have by L'hopital's rule and after simplification,
$$lim_xto1frac-n(n+1)x^n-1(1-x)-2(1-x)=lim_xto1fracn(n+1)2=fracn(n+1)2$$
From which we obtain the conclusion.

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,

$$begincasesS_1=a+b=1,\S_2=4a+2b=3endcases$$

and $$a=b=dfrac12.$$

The Lagrangian polynomial by $(0,0)$, $(1,1)$, $(2,3)$ is $dfracx^22+dfrac x2$.

$$S(n)-S(n-1)=an^2+bn-a(n-1)^2-b(n-1)=a(2n-1)+b=n$$

so that

$$a=b=dfrac12.$$

A posteriori:

$$S_n-S_n-1=fracn(n+1)2-frac(n-1)n2=n.$$

As an arithmetic progression is linear, the average term is the average of the extreme terms.

$$fracS_nn=frac1+n2$$ and $$S_n=fracn(n+1)2.$$

Let the function defined by a geometric series

$$f(x):=sum_k=0^n x^k=fracx^n+1-1x-1.$$

We differentiate on $x$,

$$f'(x)=sum_k=0^n k x^k-1=fracx^n+1-1x-1=frac(n+1)x^n(x-1)-(x^n+1-1)(x-1)^2.$$

Then with $x=1$, using L'Hospital,

$$f'(1)=sum_k=0^n k =lim_xto0fracnx^n+1-(n+1)x^n+1(x-1)^2 =lim_xto0fracn(n+1)x^n-(n+1)nx^n-12(x-1)=fracn(n+1)2.$$

The identity

$$sum_k=1^nk=fracn(n+1)2$$

is a classical result which can be easily proved by the following

Use can use the sum of AP to arrive at the same conclusion.
$a=1, d=1$
$$S_n = nover 2(2a+(n-1)d)$$
$$S_n = nover 2(2+(n-1))$$
$$S_n = nover 2(n+1)$$