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 $k = 0, 1, 2, \ldots, n - 1$


Then its closed-form expression is:

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


Proof

We have that:

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

Then:

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

So:

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

Hence the result.

$\blacksquare$


Also presented as

The Sum of Arithmetic Sequence can also be seen presented in the forms:

\(\ds \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}\) \(=\) \(\ds n a + n \dfrac 1 2 \paren {n - 1} d\)
\(\ds \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}\) \(=\) \(\ds \dfrac 1 2 n \paren {2 a + \paren {n - 1} d}\)


Some present it as:

$\ds \sum_{k \mathop = 0}^n \paren {a + k d} = a \paren {n + 1} + \dfrac 1 2 d n \paren {n + 1}$


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


Examples

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

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


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

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

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

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

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