Sum of Arithmetic Sequence

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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:

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


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:

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

So:

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

Hence the result.

$\blacksquare$


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.


Examples

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

\(\displaystyle \sum_{j \mathop = m}^n j\) \(=\) \(\displaystyle m \paren {n - m + 1} + \frac 1 2 \paren {n - m} \paren {n - m + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {n \paren {n + 1} } 2 - \frac {\paren {m - 1} m} 2\)


Sum of $i + k \paren {2 + 2 i}$

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

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

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 sequence or arithmetic-geometric sequence, 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