Arithmetic Series

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math

A type of series where the value of the next entry is equal to the current entry plus some constant.

Given the first value in the series \( a \) and the constant by which it changes \( d \), you can define the value of an element at any index in the series as

\begin{align} \label{eq:arithmetic-formula} U_n = a + (n-1)d \end{align}

All arithmetic series follow eq:arithmetic-formula.

We can define the sum of the first \( n \) elements of an arithmetic series as:

\begin{align} \label{eq:arithmetic-sum} S_n &= \frac{1}{2} n(a+l) \\

  &= \frac{1}{2} n[2a + (n-1)d]

\end{align}

The value \( l \) above refers to the last value of the series.

The origin of eq:arithmetic-sum is an interesting one and refers to the work of child prodigy Carl Friedrich Gauss. Consider an arithmetic series with \( a=1, d=1 \), that is to say \( \sum_ n \).

\begin{align} \sum_^{100} &= 1 + 2 + 3 + \ldots + 100 \
\label{eq:arithmetic-sum-proof} 2 \sum_^{100} &= 1 + 2 + 3 + \ldots + 100 \\

              &+ 100 + 99 + 98 + \ldots + 1 \\\\

2 \sum_^{100} &= 101 + 101 + 101 + \ldots + 101 \
2 \sum_^{100} &= 100 \times 101 \
2 \sum_^{100} &= \frac{100}{2} \times 101 \
\sum_^{100} &= 5050 \end{align}

In eq:arithmetic-sum-proof Gauss considers what the sum of all the elements in an arithmetic series should be. He does this by doubling the sum he's trying to find and then matching the elements at the tail of the sum with the head. All these elements sum to the same value \( a + l = 1 + 100 = 101 \) and he has \( n = 100 \) such summations therefore the total sum is \( S_n = \frac{n}{2} (a + l) \) which is \( S_{100} = 100 \frac{1+100}{2} = 550 \).

The proof derived by Gauss is general and can work with any arithmetic series.