Sum of Arithmetic Sequence

Theorem
Let $\left \langle{a_n}\right \rangle$ be an arithmetic progression defined as:
 * $a_n = a + \left({n-1}\right) d$ for $n = 1, 2, 3, \ldots$

Then its closed-form expression is:
 * $\displaystyle \sum_{j=1}^{n} \left({a + \left({j-1}\right) d}\right) = n \left({a + \frac {n - 1} 2 d}\right)$

Proof
We have that:
 * $\displaystyle \sum_{i=1}^{n}\left({a + \left({j-1}\right) d}\right) = a + \left({a+d}\right) + \left({a+2d}\right) + \cdots + \left({a+\left({n-1}\right)d}\right)$

Consider $\displaystyle 2 \sum_{i=1}^{n} \left({a + \left({j-1}\right) d}\right)$.

Then:

So:

Hence the result.

Comment
Doubt has recently been cast on the accuracy of the tale about how Gauss supposedly discovered this technique at the age of 8.