Sum of a series - Part 1

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

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

Where nn is the number of terms in the series and aa is the starting number

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

So what if we want the formula when a1a \neq 1

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

So

36=3+4+5+6+7+8=33\displaystyle \sum_{3}^{6} = 3 + 4 + 5 + 6 + 7 + 8 = 33

This sum is almost the same as

18=1+2+3+4+5+6+7+8=36\displaystyle \sum_{1}^{8} = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36

The second sum can be worked out using the formula

18=8(8+1)2=36\displaystyle \sum_{1}^{8} = \frac {8(8 + 1)}{2} = 36

And the first one is just

36=18(1+2)\displaystyle \sum_{3}^{6} = \displaystyle \sum_{1}^{8} - (1 + 2)

Since (1+2)(1 + 2) can be represented as 12\displaystyle \sum_{1}^{2}, we can represent our equation as

36=1812\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

36=8(8+1)22(2+1)2=363=33\displaystyle \sum_{3}^{6} = \frac {8(8 + 1)}{2} - \frac {2(2 + 1)}{2} = 36 - 3 = 33

Rewriting this using aa and nn gives us

an=1n+(a1)1a1\displaystyle \sum_{a}^{n} = \displaystyle \sum_{1}^{n + (a - 1)} - \displaystyle \sum_{1}^{a - 1}

an=(n+(a1))((n+(a1))+1)2(a1)((a1)+1)2\displaystyle \sum_{a}^{n} = \frac {(n + (a - 1))((n + (a - 1)) + 1)}{2} - \frac {(a - 1)((a - 1) + 1)}{2}

an=(n+a)(n+a1)2a2a2\displaystyle \sum_{a}^{n} = \frac {(n + a)(n + a - 1)}{2} - \frac {a^2 - a}{2}

Lets test it on our example

36=(6+3)(6+31)23232=363=33\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=3a = 3 and n=6n = 6), feel free to test it for yourself.

Note by Jack Rawlin
4 years, 9 months ago

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