Proof that $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ Why is $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$ $\space$ ?

Gathering as many proofs as we can? Write the series recursively:
$$S(n) = S(n - 1) + n \tag{1}$$
Substitute $n \to n+1$ :
$$S(n + 1) = S(n) + n + 1\tag{2}$$
Equation (2) subtract Equation (1):
$$S(n+1) - S(n) = S(n) + 1 - S(n - 1) \tag{3}$$
And write it up:
$$\begin{cases} 
S(n+1) &= 2S(n) -S(n-1) + 1 \\
S(n) &= S(n) \end{cases} \tag{4}$$
Which can now be written in matrix form:
$$
\begin{bmatrix} S(n+1) \\ S(n) \end{bmatrix} =
\begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}
\begin{bmatrix} S(n) \\ S(n-1) \end{bmatrix} + 
\begin{bmatrix} 1 \\ 0 \end{bmatrix}
\tag{5}$$
And then converting the affine equation (5) to the linear equation (6):
$$
\begin{bmatrix} S(n+1) \\ S(n+0) \\ 1\end{bmatrix} =
\begin{bmatrix} 2 & -1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}
\begin{bmatrix} S(n) \\ S(n-1) \\ 1\end{bmatrix}
\tag{6}$$
And closing the equation:
$$
\begin{bmatrix} S(n+1) \\ S(n) \\ 1\end{bmatrix} =
\begin{bmatrix} 2 & -1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}^n
\begin{bmatrix} S(1) \\ S(0) \\ 1\end{bmatrix}
\tag{6}$$
Then finding the Jordan form of the 3x3 matrix:
$$\begin{align}
\begin{bmatrix} S(n+1) \\ S(n) \\ 1\end{bmatrix} &=
\left(\begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}
\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}
\begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}^{-1}\right)^n
\begin{bmatrix} S(1) \\ S(0) \\ 1\end{bmatrix}
\\ &=
\begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}
\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}^n
\begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}^{-1}
\begin{bmatrix} S(1) \\ S(0) \\ 1\end{bmatrix}
\\ &=
\begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}
\begin{bmatrix} 1 & \binom{n}{1} & \binom{n-1}{2} \\ 0 & 1 & \binom{n}{1} \\ 0 & 0 & 1\end{bmatrix}
\begin{bmatrix} 0 & 1 & 0 \\ 1 & -1 & 0 \\ 0 & 0 & 1\end{bmatrix}
\begin{bmatrix} S(1) \\ S(0) \\ 1\end{bmatrix}
\end{align}
\tag{7}$$
And multiplying the matrices out:
$$\begin{bmatrix} S(n+1) \\ S(n) \\ 1\end{bmatrix} =
\begin{bmatrix}
\frac{ (2n+2)S(1) - 2nS(0) + {n}^{2} + n}{2} \\
\frac{ 2nS(1) + (2 - 2n)S(0)+{n}^{2}-n}{2}    \\
1
\end{bmatrix} \tag{8}$$
And given that $S(0) = 0$ and $S(1) = 1$, we get that:
$$S(n) = \frac{n^2 + n}{2} \tag{9}$$
