A calculus of the absurd

3.3 Arithmetic series

An arithmetic sequence goes something like

\begin{equation} a, a + d, a + 2d, a + 3d, ..., a + (n-1)d \end{equation}

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

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

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

\begin{equation} 1 + 2 + 3 + ... + \rbrackets {n-1} \end{equation}

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 \(n=5\)

\begin{equation} 1 + 2 + 3 + 4 + 5 = 15 \end{equation}

Or when \(n=6\)

\begin{equation} 1 + 2 + 3 + 4 + 5 + 6 = 21 \end{equation}

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

\begin{equation} 1 + 2 + 3 + ... + \rbrackets {n-3} + \rbrackets {n-2} + \rbrackets {n-1} \end{equation}

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

\begin{equation} \frac {1}{2} (n-1) (n) \end{equation}

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 3.25 is also what is given in the formula booklet.