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:
 * $$\sum_{j=1}^{n} \left({a + \left({j-1}\right) d}\right) = n \left({a + \frac {n - 1} 2 d}\right)$$.

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

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