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Finding Sums of Finite Arithmetic Series

Concept

Series

A series is a summation of the terms in a sequence. For the sequence $2,$ $4,$ $6,$ $8,$ the series can be written as $2+4+6+8.$ If the terms are given by a rule, it's usually more compact to write it using sigma notation. The sequence above is described by the explicit rule $a_n=2n.$ The series can be written as $\sum_{n=1}^4 2n.$ Here, by substituting $n$ with integers $1$ (under $\Sigma$) through $4$ (above $\Sigma$) the individual terms of the series are obtained, and the sigma itself indicates that between each term is a plus sign. This alternative notation is useful if the series contains a large number of terms, and especially useful when the sequence is infinite. Then, the infinity symbol is written above the sigma.

$\sum_{n=1}^{\infty} 2n$
Concept

Arithmetic Series

The summation of the terms in an arithmetic sequence is called an arithmetic series. If the sequence is short enough, such as $1,3,5,7,$ it's straightforward to calculate the sum: $s=1+3+5+7=16.$

However, if the sequence is longer, it can be tedious to add the terms by hand. In that case, the formula for an arithmetic sum can be used.
Rule

Sum of an Arithmetic Series

For a finite arithmetic sequence given by $a_n=a_1+(n−1)d,$ where $a_1$ is the first term, $d$ is the common difference, and $n$ is the number of terms, the sum of all terms, $S_n,$ can be calculated using the following formula.

$S_n=\dfrac{n(a_1+a_n)}{2}$

To explain why the sum is determined this way, consider the arithmetic sum $1+2+3+\ldots+9.$ Instead of adding all terms in one go, note that the first and last term can be rewritten as $2$ fives. The second and second to last term can be rewritten in the same way. In fact, all opposite pairs can be rewritten as $2$ fives.
Rewrite pairs

Now, there are $9$ fives, and the sum is $9\cdot5.$ Here, $5$ is the mean of the first and last term, which means that it can be written as $\dfrac{a_1+a_n}{2}.$ This is multiplied by $9,$ which is the number of terms, $n,$ leading to the formula

$S_n=n\cdot\dfrac{a_1+a_n}{2} = \dfrac{n(a_1+a_n)}{2}.$
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Exercise

Determine the sum of all positive integers between $1$ and $1000.$

Show Solution
Solution
The sequence from $1$ through $1000$ is arithmetic since the difference between each term is constant, $1.$ The sum of the series is then $S_n=\dfrac{n(a_1+a_n)}{2}.$ The first term, $a_1,$ is $1$ and the last, $a_n,$ is $1000.$ The number of terms is $n=1000.$
$S_n=\dfrac{n(a_1+a_n)}{2}$
$S_{{\color{#0000FF}{1000}}}=\dfrac{{\color{#0000FF}{1000}}(a_1+a_{{\color{#0000FF}{1000}}})}{2}$
$S_{1000}=500(a_1+a_{1000})$
$S_{1000}=500({\color{#0000FF}{1}}+{\color{#009600}{1000}})$
$S_{1000}=500+500\,000$
$S_{1000}=500\,500$
The sum of the all positive integers up to $1000$ is $500\,500.$