# 3.3 Arithmetic series

An arithmetic sequence goes something like

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

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

To find this, we can get creative in how we group the terms in the sum. We can
think about some smaller sequences,
^{3}^{3}
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 $n=5$

Or when $n=6$

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

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

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.