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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, 3 months ago

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