Sum of Arithmetic Sequence

Theorem
Let $\left \langle{a_n}\right \rangle$ be an arithmetic progression defined as:
 * $a_n = a + n d$ for $n = 0, 1, 2, \ldots$

Then its closed-form expression is:
 * $\displaystyle \sum_{j \mathop = 0}^{n - 1} \left({a + j d}\right) = n \left({a + \frac {n - 1} 2 d}\right)$

Proof
We have that:


 * $\displaystyle \sum_{j \mathop = 0}^{n - 1} \left({a + j d}\right) = a + \left({a + d}\right) + \left({a + 2 d}\right) + \cdots + \left({a + \left({n - 1}\right) d}\right)$

Then:

So:

Hence the result.

Linguistic Note
In this context, the word arithmetic is pronounced with the stress on the first and third syllables: a-rith-me-tic, rather than on the second syllable: a-rith-me-tic.

This is because the word is being used in its adjectival form.

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