Sum of Arithmetic Sequence

Theorem
Let $\sequence {a_k}$ be an arithmetic sequence defined as:
 * $a_k = a + k d$ for $n = 0, 1, 2, \ldots, n - 1$

Then its closed-form expression is:

Proof
We have that:


 * $\displaystyle \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} = a + \paren {a + d} + \paren {a + 2 d} + \dotsb + \paren {a + \paren {n - 1} d}$

Then:

So:

Hence the result.

Also presented as
The sum can also be seen presented in the forms:


 * $n a + n \dfrac 1 2 \paren {n - 1} d$


 * $\dfrac 1 2 n \paren {2 a + \paren {n - 1} d}$

The reality is that this is a messy result that cannot be presented elegantly.