On the previous note in this series we learnt / revised that

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

That's the formula when the difference ($d$) is $1$ so what would the formula be if $d \neq 1$

Let's denote the whole thing as $\displaystyle \sum_{a}^{n} d$

So let's say that $a = 3$, $n = 6$ and $d = 2$ what equation would we get from that.

$\displaystyle \sum_{3}^{6} 2 = 3 + 5 + 7 + 9 + 11 + 13 = 48$

We're going to have to use a different method to last time to solve this.

Since $a = 3$ and $d = 2$ we can put those in to get

$\displaystyle \sum_{a}^{6} d = a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) + (a + 5d)$

That can be written as

$\displaystyle \sum_{a}^{6} d = 6a + (1 + 2 + 3 + 4 + 5)d$

$\displaystyle \sum_{a}^{n} d = na + \frac {dn(n-1)}{2}$

This is basically a simplified version of the previous equation with a $d$ added in to account for the difference. This formula however is still flawed as it can only handle a constant variable for $d$.

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