The running totals of 1,2,3 ... are called the
triangular numbers. We will show that $1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$
Proof-1 Gauss's proof
( Gauss supposedly came up with this proof when he was 8 years old. On
this page you will find more than 100 different tellings of this story. )
$s = 1 + 2 + 3 + \cdots + n$
$\underline{s = n + (n-1) + (n-2) + \cdots + 1}$
$2s = (n + 1) + ((n-1)+2) + ((n-2)+3) + \cdots + (1+n) \Leftrightarrow $
$2s = n \cdot (n + 1) \Leftrightarrow $
$s = \frac{n(n+1)}{2}$
Proof-2 By induction
Let $S=\left\{ n \in \mathbf{N} | \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \right\}$
Clearly $1 \in S$
Assume, $n \in S$:
$\sum_{k=1}^{n+1} k = \frac{n(n+1)}{2} + (n+1) = \frac{(n+1)(n+2)}{2}$, or $n \in S \Rightarrow n+1 \in S$
Now, since $(S \subset \mathbf{N} \wedge 1 \in S \wedge n \in S \Rightarrow n+1 \in S ) \Rightarrow S=\mathbf{N}.$
Proof-3 With the Pascal Triangle
Because $n^k$ is in the PT for any $k \in \mathbf{N}$, sums of polynomials with integer coefficients can be read from the PT. ( $n={n \choose 1}$, $n^2={n \choose 1} + 2{n \choose 2}$, and so forth. )
$\underline{n}$
0: 1
1: 1 - 1
2: 1 - 2 - 1
3: 1 - 3 - 3 - 1
4: 1 - 4 - 6 - 4 - 1
5: 1 - 5 - 10-10 - 5 - 1
From $n$ we seek the second column, one row down or ${n+1 \choose 2}= \frac{n(n+1)}{2}$
Proof-4 Geometric
Look at the pattern
X
and
X---Y
X
X-X
and
X---Y-Y
X-X---Y
X
X-X
X-X-X
and
X---Y-Y-Y
X-X---Y-Y
X-X-X---Y
The number of X's and Y's are equal. The triangle X-pattern with base of n X's is replaced by a rectangular shape of n+1 by n X's OR Y's.
Proof-5 With Discrete Calculus
The discrete analog of solving a differential equation.
$\Delta f(n) = n+1$
$\Sigma \Delta f(n) = \Sigma (n+1) $
$f(n) = \frac{1}{2}(n+1)^{\underline{2}} + C$
$f(n) = \frac{1}{2}(n+1)n + C$
Since $f(1) = 1, C=0$
$f(n) = \frac{n(n+1)}{2}$
