Arithmetic Series and Sigma Notation

Quick answer

An arithmetic series adds the terms of an arithmetic sequence. Pairing the first term with the last, the second with the second-to-last, and so on gives pairs with equal sums, so Sₙ = n(a₁ + aₙ)/2: the number of terms times the average of the first and last. Young Carl Friedrich Gauss is said to have used this trick to add 1 through 100 in his head. Sigma notation writes a sum compactly: the sum of aₖ from k = 1 to n means a₁ + a₂ + ⋯ + aₙ.

What you'll learn

Find the sum of an arithmetic series with the pairing formula
Count the terms of an arithmetic sequence correctly
Read and write sums in sigma notation
Solve for the number of terms that gives a given sum

Adding an arithmetic sequence

An arithmetic series is the sum of the terms of an arithmetic sequence, such as $1 + 2 + 3 + \cdots + 100$ . Adding the terms one by one works, slowly. The story goes that the young Carl Friedrich Gauss, told to add $1$ through $100$ , wrote the sum twice, once backward:

\begin{aligned} S &= 1 + 2 + 3 + \cdots + 100\\ S &= 100 + 99 + 98 + \cdots + 1 \end{aligned}

Adding column by column gives $100$ pairs, each summing to $101$ . So $2S = 100 \cdot 101$ and $S = 5050$ .

Why the pairing always works

In any arithmetic sequence, moving one step in from the front adds $d$ , and moving one step in from the back subtracts $d$ . So the second and second-to-last terms have the same sum as the first and last, and so does every pair after them. Paired from the outside in, the terms make equal pairs. So the sum is the number of terms times the average of the first and last:

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

Two copies of 1 + 2 + 3 + 4 + 5 make a rectangle

In the picture, two staircases of $1 + 2 + 3 + 4 + 5$ fill a $5 \times 6$ rectangle, so each holds $\tfrac{30}{2} = 15$ . With $a_n = a_1 + (n - 1)d$ , the formula can also be written $S_n = \tfrac{n}{2}\big(2a_1 + (n - 1)d\big)$ .

Sigma notation

The Greek capital sigma, $\Sigma$ , is shorthand for a sum:

\sum_{k=1}^{n} a_k = a_1 + a_2 + \cdots + a_n

The index $k$ runs through every whole number from the bottom value to the top. Sums of this kind split and scale like ordinary addition:

\sum_{k=1}^{n} (a_k + b_k) = \sum_{k=1}^{n} a_k + \sum_{k=1}^{n} b_k \qquad \sum_{k=1}^{n} c\,a_k = c\sum_{k=1}^{n} a_k \qquad \sum_{k=1}^{n} k = \frac{n(n + 1)}{2}

Worked examples

Common mistakes

Practice problems

Find $2 + 4 + 6 + \cdots + 50$ .

Answer

$650$

Full solution

There are $25$ terms, so the sum is $\tfrac{25(2 + 50)}{2}$ .
Find $\displaystyle\sum_{k=1}^{10} k$ .

Answer

$55$

Full solution

$\tfrac{10 \cdot 11}{2}$ .
Find $\displaystyle\sum_{k=1}^{12} (2k - 1)$ .

Answer

$144$

Full solution

The first $12$ odd numbers add to $12^2$ .
Find $\displaystyle\sum_{k=1}^{20} (4k + 3)$ .

Answer

$900$

Full solution

$4 \cdot \tfrac{20 \cdot 21}{2} + 3 \cdot 20 = 840 + 60$ .
Find $7 + 10 + 13 + \cdots + 100$ .

Answer

$1712$

Full solution

$\tfrac{100 - 7}{3} + 1 = 32$ terms, so the sum is $\tfrac{32(7 + 100)}{2} = 16 \cdot 107$ .
Write $3 + 6 + 9 + \cdots + 60$ in sigma notation and find its value.

Answer

$\sum_{k=1}^{20} 3k = 630$

Full solution

The terms are $3k$ for $k = 1$ to $20$ , and $3 \cdot \tfrac{20 \cdot 21}{2} = 630$ .
A pile of logs has $20$ logs in the bottom row, $19$ in the next, and so on up to $1$ on top. How many logs are there?

Answer

$210$

Full solution

$1 + 2 + \cdots + 20 = \tfrac{20 \cdot 21}{2}$ .
You save $5$ dollars the first week, $7$ the second, and $2$ dollars more each week after. How much have you saved after $26$ weeks?

Answer

$780$ dollars

Full solution

The last week’s amount is $5 + 25 \cdot 2 = 55$ , so the total is $\tfrac{26(5 + 55)}{2}$ .
How many terms of $2 + 5 + 8 + \cdots$ add up to $155$ ?

Answer

$10$

Full solution

The $n$ th term is $3n - 1$ , so $S_n = \tfrac{n(3n + 1)}{2} = 155$ . Then $3n^2 + n - 310 = 0$ , and the quadratic formula gives $n = \tfrac{-1 + 61}{6} = 10$ .
A student counts the terms of $7 + 10 + 13 + \cdots + 100$ as $\tfrac{100 - 7}{3} = 31$ and gets a sum of $1658.5$ . What went wrong?

Hint

A sum of whole numbers cannot end in $.5$ . How many terms are in $7, 10, 13$ ?

Answer

The student counted the gaps, not the terms. There are $32$ terms, and the sum is $1712$ .

Full solution

$\tfrac{100 - 7}{3} = 31$ is the number of steps of $3$ between the first and last terms. Three terms, $7, 10, 13$ , have only two steps between them, so the count of terms is always one more than the count of steps.

Frequently asked questions

What is the formula for the sum of an arithmetic series?

Sₙ = n(a₁ + aₙ)/2: the number of terms times the average of the first and last terms. Equivalently, Sₙ = n(2a₁ + (n − 1)d)/2.

What is 1 + 2 + 3 + ⋯ + n?

n(n + 1)/2. For example, 1 + 2 + ⋯ + 100 = 100 · 101 / 2 = 5050.

How do I count the terms from 7 to 100 going up by 3?

Divide the distance by the step and add 1: (100 − 7)/3 + 1 = 32. The +1 counts the first term.

What does sigma notation mean?

The Greek letter Σ means add up. The sum of aₖ from k = 1 to n is a₁ + a₂ + ⋯ + aₙ, one term for each value of k.

What is the sum of the first n odd numbers?

n². The first term is 1 and the last is 2n − 1, so the sum is n(1 + 2n − 1)/2 = n².

What to learn next

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

CCSS.MATH.CONTENT.HSF.BF.A.2Building FunctionsWrite arithmetic and geometric sequences both recursively and with an explicit formula, use them to model situations, and translate between the two forms.

For teachersPrint problems onlyTick to print a blank worksheet; leave it clear to print the lesson with the full answer key.