Sum of Arithmetic Progression

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Theorem

Let $\sequence {a_k}$ be an arithmetic progression defined as:

$a_k = a + k d$ for $n = 0, 1, 2, \ldots, n - 1$


Then its closed-form expression is:

\(\displaystyle \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}\) \(=\) \(\displaystyle n \paren {a + \frac {n - 1} 2 d}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {n \paren {a + l} } 2\) $\quad$ where $l$ is the last term of $\sequence {a_k}$ $\quad$


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:

\(\displaystyle 2 \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}\) \(=\) \(\displaystyle 2 \paren {a + \paren {a + d} + \paren {a + 2 d} + \dotsb + \paren {a + \paren {n - 1} d} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \paren {a + \paren {a + d} + \dotsb + \paren {a + \paren {n - 1} d} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\, \displaystyle + \, \) \(\displaystyle \paren {\paren {a + \paren {n - 1} d} + \paren {a + \paren {n - 2} d} + \dotsb + \paren {a + d} + a}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \paren {2 a + \paren {n - 1} d}_1 + \paren {2 a + \paren {n - 1} d}_2 + \dotsb + \paren {2 a + \paren {n - 1} d}_n\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle n \paren {2 a + \paren {n - 1} d}\) $\quad$ $\quad$

So:

\(\displaystyle 2 \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}\) \(=\) \(\displaystyle n \paren {2 a + \paren {n - 1} d}\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}\) \(=\) \(\displaystyle \frac {n \paren {2 a + \paren {n - 1} d} } 2\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {n \paren {a + l} } 2\) $\quad$ Definition of Last Term $l$ $\quad$

Hence the result.

$\blacksquare$


Examples

Sum of $j$ from $m$ to $n$

\(\displaystyle \sum_{j \mathop = m}^n j\) \(=\) \(\displaystyle m \left({n - m + 1}\right) + \frac 1 2 \left({n - m}\right) \left({n - m + 1}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {n \left({n + 1}\right)} 2 - \frac {\left({m - 1}\right) m} 2\) $\quad$ $\quad$


Sum of $i + k \left({2 + 2 i}\right)$

Let $A_n$ be the arithmetic progression of $n$ terms defined as:

\(\displaystyle A_n\) \(=\) \(\displaystyle \sum_{k \mathop = 0}^{n - 1} \paren {a_0 + \paren {2 + 2 i} k}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle i + \paren {2 + 3 i} + \paren {4 + 5 i} + \paren {6 + 7 i} + \dotsb + \paren {2 n - 2 + \paren {2 n - 1} i}\) $\quad$ $\quad$

Then:

$A_n = n \paren {n - 1} + n^2 i$


Historical Note

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


Linguistic Note

In the context of an arithmetic progression or arithmetic-geometric progression, 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.


Sources