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Sum of a series - Part 1

So we know the general formula for a sum of an arithmetic series is

\[\displaystyle \sum_{a}^{n} = \frac {n(n + 1)}{2}\]

Where \(n\) is the number of terms in the series and \(a\) is the starting number

The problem with this formula though is that it only handles when \(a=1\)

So what if we want the formula when \(a \neq 1\)

Let's say that for now \(a = 3\) and \(n = 6\) so we have an answer to check our equation with.

So

\[\displaystyle \sum_{3}^{6} = 3 + 4 + 5 + 6 + 7 + 8 = 33\]

This sum is almost the same as

\[\displaystyle \sum_{1}^{8} = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36\]

The second sum can be worked out using the formula

\[\displaystyle \sum_{1}^{8} = \frac {8(8 + 1)}{2} = 36\]

And the first one is just

\[\displaystyle \sum_{3}^{6} = \displaystyle \sum_{1}^{8} - (1 + 2)\]

Since \((1 + 2)\) can be represented as \(\displaystyle \sum_{1}^{2}\), we can represent our equation as

\[\displaystyle \sum_{3}^{6} = \displaystyle \sum_{1}^{8} - \displaystyle \sum_{1}^{2}\]

And since both of those can be represented using the formula we can rewrite it as

\[\displaystyle \sum_{3}^{6} = \frac {8(8 + 1)}{2} - \frac {2(2 + 1)}{2} = 36 - 3 = 33\]

Rewriting this using \(a\) and \(n\) gives us

\[\displaystyle \sum_{a}^{n} = \displaystyle \sum_{1}^{n + (a - 1)} - \displaystyle \sum_{1}^{a - 1}\]

\[\displaystyle \sum_{a}^{n} = \frac {(n + (a - 1))((n + (a - 1)) + 1)}{2} - \frac {(a - 1)((a - 1) + 1)}{2}\]

\[\displaystyle \sum_{a}^{n} = \frac {(n + a)(n + a - 1)}{2} - \frac {a^2 - a}{2}\]

Lets test it on our example

\[\displaystyle \sum_{3}^{6} = \frac {(6 + 3)(6 + 3 - 1)}{2} - \frac {3^2 - 3}{2} = 36 - 3 = 33\]

So it works (at least for \(a = 3\) and \(n = 6\)), feel free to test it for yourself.

Note by Jack Rawlin
2 years, 5 months ago

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