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

I didn't see an answer using the method that I used, so I'm posting an answer here to 'spread the knowledge!'
This can be applied to higher powers such $1^2+2^2+\cdots n^2$ or $1^3+2^3+\cdots n^3$.


Solving by the use of Indeterminate Coefficients:

Assume the series$$1+2+3+4+5\ldots+n\tag1$$
Is equal to the infinite series$$1+2+3+4+5+\ldots+n=A+Bn+Cn^2+Dn^3+En^4+\ldots\&c\tag2$$
If we 'replace' $n$ with $n+1$, we get$$1+2+3+4+\ldots+(n+1)=A+B(n+1)+C(n+1)^2+\ldots\&c\tag3$$
And subtracting $(3)-(2)$, gives$$\begin{align*} & n+1=B+C(2n+1)\tag4\\n & +1=2Cn+(B+C)\tag5\end{align*}$$
Therefore, $C=\dfrac 12,B=\dfrac 12,A=0$ and $(1)$ becomes$$1+2+3+4+5\ldots+n=\dfrac n2+\dfrac {n^2}2=\dfrac {n(n+1)}{2}$$
