3.3 Arithmetic series ‣ Chapter 3 Sequences and series ‣ Part I A Level mathematics (and further mathematics) ‣ A calculus of the absurd‣ Chapter 3 Sequences and series ‣ Part I A Level mathematics (and further mathematics) ‣ A calculus of the absurd

An arithmetic sequence goes something like

a,a+d,a+2d,a+3d,...,a+(n-1)d

(3.26)

An arithmetic series is the sum of this sequence. If we want to find the sum of this series, it goes something like

S_{n}=\left[a\right]+\left[a+d\right]+\left[a+2d\right]+...+\left[a+(n-1)d\right]

(3.27)

If we add all the $a$ s and the $d$ s seperately, we get that

	$\displaystyle S_{n}$	$\displaystyle=\left[n\cdot a\right]+\left[d+2d+3d+...+(n-1)d\right]$		(3.28)
		$\displaystyle=\left[n\cdot a\right]+\left[\left(1+2+3+...+\left(n-1\right)% \right)d\right]$		(3.29)

Here, the question becomes one of the value of

1+2+3+...+\left(n-1\right)

(3.30)

To find this, we can get creative in how we group the terms in the sum. We can think about some smaller sequences, ³³ 3 A useful heuristic when working with sums of sequences is to first consider the sum of a few terms and then try to generate to $n$ terms for example when $n=5$

1+2+3+4+5=15

(3.31)

Or when $n=6$

1+2+3+4+5+6=21

(3.32)

In general, the "trick" here is to group the terms quite imaginatively. In each case, we can add up the two terms on opposite sides of the sequence (e.g. $1+6=7$ , as is $2+3$ and as is $3+4$ ).

Taking the general case, we have

1+2+3+...+\left(n-3\right)+\left(n-2\right)+\left(n-1\right)

(3.33)

When we add up the terms on opposite ends we get that

$\displaystyle 1+\left(n-1\right)$	$\displaystyle=n$	(3.34)
$\displaystyle 2+\left(n-2\right)$	$\displaystyle=n$	(3.35)
$\displaystyle 3+\left(n-3\right)$	$\displaystyle=n$	(3.36)
$\displaystyle k+\left(n-k\right)$	$\displaystyle=n$	(3.37)

Because every two terms sum to $n$ , for the sum from $1$ to $n-1$ we have

\frac{1}{2}(n-1)(n)

(3.38)

This is because we have $n-1$ total items in our sequence, and every two terms sum to $n$ , so the total sum is half the number of terms (the number of paired items adding together to give $n$ ).

Therefore, returning to our arithmetic series, overall

$\displaystyle S_{n}$	$\displaystyle=n\cdot a+\frac{1}{2}(n-1)\cdot n\cdot d$	(3.39)
	$\displaystyle=n\left[a+\frac{1}{2}(n-1)d\right]$	(3.40)
	$\displaystyle=\frac{n}{2}\left[a+\frac{1}{2}(n-1)d\right]$	(3.41)

Equation 3.41 is also what is given in the formula booklet.