Sum of a series - Part 2

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 \(d\).

Note by Jack Rawlin
3 years, 6 months ago

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