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

You can also prove it by induction, which is nice and easy, although it doesn't give much intuition as to why it works.
It works for $1$ since $\frac{1\times 2}{2} = 1$.
Let it work for $n$.
$$1 + 2 + 3 + \dots + n + (n + 1) = \frac{n(n+1)}{2} + (n + 1) = \frac{n(n+1) + 2(n+1)}{2} = \frac{(n+1)(n+2)}{2}.$$
Therefore, if it works for $n$, it works for $n + 1$. Hence, it works for all natural numbers.
For the record, you can see this by applying the formula for the sum of an arithmetic progression (a sequence formed by constantly adding a rate to each term, in this case $1$). The formula is reached pretty much using the method outlined by Carl earlier in this post. Here it is, in all its glory:
$$S_n = \frac{(a_1 + a_n) * n}{2}$$
($a_1$ = first term, $a_n$ = last term, $n$ = number of terms being added).
