# A calculus of the absurd

#### 3.3 Arithmetic series

An arithmetic sequence goes something like

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

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

$$S_n = \sbrackets {a} + \sbrackets {a + d} + \sbrackets {a + 2d} + ... + \sbrackets {a + (n-1)d}$$

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

\begin{align} S_n & = \sbrackets {n \cdot a} + \sbrackets { d + 2d + 3d + ... + (n-1)d } \\ & = \sbrackets {n \cdot a} + \sbrackets { \rbrackets { 1 + 2 + 3 + ... + \rbrackets {n-1} }d } \end{align}

Here, the question becomes one of the value of

$$1 + 2 + 3 + ... + \rbrackets {n-1}$$

To find this, we can get creative in how we group the terms in the sum. We can think about some smaller sequences, 33 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 $$n=5$$

$$1 + 2 + 3 + 4 + 5 = 15$$

Or when $$n=6$$

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

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

$$1 + 2 + 3 + ... + \rbrackets {n-3} + \rbrackets {n-2} + \rbrackets {n-1}$$

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

\begin{align} & 1 + \rbrackets {n - 1} & = n \\ & 2 + \rbrackets {n - 2} & = n \\ & 3 + \rbrackets {n - 3} & = n \\ & k + \rbrackets {n - k} & = n \end{align}

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

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

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

\begin{align} S_n & = n\cdot a + \frac {1}{2}(n-1)\cdot n \cdot d \\ & = n \sbrackets {a + \frac {1}{2}(n-1)d} \\ & = \frac {n}{2} \sbrackets {a + \frac {1}{2}(n-1)d} \label {in formula sheet} \end{align}

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