Sum of a series - Part 2 - Jack Rawlin

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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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
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Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$