Sum of Arithmetic Progression

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 2 \sum_{i=1}^n \left({a + \left({j-1}\right) d}\right)\)	\(=\)	\(\displaystyle 2 a + \left({a+d}\right) + \left({a+2d}\right) + \cdots + \left({a+\left({n-1}\right)d}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({a + \left({a+d}\right) + \left({a+2d}\right) + \cdots + \left({a+\left({n-1}\right)d}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(+\)	\(\displaystyle \left({\left({a+\left({n-1}\right)d}\right) + \left({a+\left({n-2}\right)d}\right) + \cdots + \left({a+d}\right) + a}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({2 a + \left({n-1}\right) d}\right)_1 + \left({2 a + \left({n-1}\right) d}\right)_2 + \cdots + \left({2 a + \left({n-1}\right) d}\right)_n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle n \left({2 a + \left({n-1}\right) d}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

So:

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

Hence the result.

$\blacksquare$

Comment

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