<s>[INST] Proof that $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ Why is $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$ $\space$ ?
 [/INST] 
Above is an image representing Pascals Triangle. What I want to draw attention to is the hockey stick formation, particularly, the blue hockey stick. Notice how the entries in the stick of the blue hockey stick are in arithmetic progression, and that the entry in the blade represents the sum of the entries in the stick.

To prove this inductively we have as a bootstrap condition
$$1=\frac{1(1+1)}{2}=\binom{1+1}{2} = \sum\limits_{i=1}^1\binom{i}{1}=1$$
and for the general case
$$\begin{array}{lll}
\sum\limits_{i=1}^{n+1}&=&(n+1)+\sum\limits_{i=1}^{n}i\\
&=&\binom{n+1}{1}+\binom{n+1}{2}\\
&=&\binom{n+2}{2}\\
&=&\binom{(n+1)+1}{2}\\
&=&\frac{(n+1)((n+1)+1)}{2}
\end{array}$$
Of course, we assumed that $\binom{n}{k}+\binom{n}{k+1} = \binom{n+1}{k+1}$ holds.
$$\begin{array}{lll}
\binom{n}{k}+\binom{n}{k+1}&=&\frac{n!}{k!(n-k)!} + \frac{n!}{(k+1)!(n-(k+1))!}\\
&=&\frac{n!(k+1)}{k!(k+1)(n-k)!} + \frac{n!(n-k)}{(k+1)!(n-(k+1))!(n-k)}\\
&=&\frac{n!k+n!+n!n-n!k}{(k+1)!(n-k)!}\\
&=&\frac{n!+n!n}{(k+1)!((n+1)-(k+1))!}\\
&=&\frac{n!(n+1)}{(k+1)!((n+1)-(k+1))!}\\
&=&\frac{(n+1)!}{(k+1)!((n+1)-(k+1))!}\\
&=&\binom{n+1}{k+1}\\
\end{array}$$
</s>