Sum of a series - Part 2

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

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

That's the formula when the difference (dd) is 11 so what would the formula be if d1d \neq 1

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

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

362=3+5+7+9+11+13=48\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=3a = 3 and d=2d = 2 we can put those in to get

a6d=a+(a+d)+(a+2d)+(a+3d)+(a+4d)+(a+5d)\displaystyle \sum_{a}^{6} d = a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) + (a + 5d)

That can be written as

a6d=6a+(1+2+3+4+5)d\displaystyle \sum_{a}^{6} d = 6a + (1 + 2 + 3 + 4 + 5)d

and=na+dn(n1)2\displaystyle \sum_{a}^{n} d = na + \frac {dn(n-1)}{2}

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

Note by Jack Rawlin
4 years, 9 months ago

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