Algebra II/Functions/Arithmetic Series and Sigma Notation

Algebra II Regents in 15 days

Algebra II · Lesson 1

Arithmetic Series and Sigma Notation

Adding up an arithmetic sequence — and a shortcut that saves you from doing it term by term.

In 1787, a teacher gave his class of young students a punishment problem: add every integer from 1 to 100. The teacher expected it to take the class the rest of the period. One student, Carl Friedrich Gauss, wrote the answer in seconds. He had noticed that if you pair the first term with the last, the second with the second-to-last, and so on, every pair sums to 101. There are 50 such pairs. So the total is $50 \times 101 = 5050$ . Gauss had just discovered, at age 10, the formula for an arithmetic series.

An arithmetic series is what you get when you add the terms of an arithmetic sequence. The sequence $3, 7, 11, 15$ becomes the series $3 + 7 + 11 + 15$ . The difference between the two ideas is the plus signs: a sequence lists values, a series adds them.

Before deriving the formula, here is the notation that makes series easier to write. The capital Greek letter sigma, $\Sigma$ , means "sum." When you write

\sum_,^, a_k

you mean: evaluate $a_k$ at every integer value of $k$ from 1 to $n$ , and add the results. The number at the bottom of the sigma is where you start. The number at the top is where you stop. So

\sum_,^, (2k + 1)

means plug in $k = 1, 2, 3, 4$ into the expression $2k+1$ and add everything up.

Expanding a sigma notation expression

\sum_{k=1}^{4} (2k + 1) = (2(1)+1) + (2(2)+1) + (2(3)+1) + (2(4)+1)

Replace k with 1, then 2, then 3, then 4. Write out each term before adding.

= 3 + 5 + 7 + 9

Evaluate each expression: 2(1)+1=3, 2(2)+1=5, and so on.

= 24 \checkmark

Add them up. This is an arithmetic series with first term 3, last term 9, and 4 terms.

Now for the formula Gauss found. Take any arithmetic series with $n$ terms, first term $a_1$ , and last term $a_n$ . Write the sum forward, then write it backward, and add the two rows together.

Forward: KaTeX can only parse string typed expression

Backward: KaTeX can only parse string typed expression

Every vertical pair sums to $a_1 + a_n$ , and there are $n$ pairs. So the two rows together equal $n(a_1 + a_n)$ . But that is twice the sum you want, so divide by 2.

S_n = \frac,(a_1 + a_n)

This is the formula for the sum of a finite arithmetic series. You need three things: the number of terms $n$ , the first term $a_1$ , and the last term $a_n$ . If you know any two of those you can usually find the third using what you already know about arithmetic sequences.

Sum of the first 20 terms: 5 + 8 + 11 + ...

a_1 = 5, \quad d = 3, \quad n = 20

Read the sequence: it starts at 5 and goes up by 3 each time.

a_{20} = 5 + (20-1)(3)

Use the arithmetic sequence formula to find the last term you need.

a_{20} = 5 + 57 = 62

19 steps of 3 past the starting value.

S_{20} = \frac{20}{2}(5 + 62)

Plug into the series formula: n=20, first term=5, last term=62.

S_{20} = 10 \cdot 67 = 670 \checkmark

Half of 20 is 10. Multiply by the sum of the endpoints.

Evaluating a series written in sigma notation

\sum_{k=1}^{50} (3k - 1)

This is a sum of 50 terms. The variable k goes from 1 to 50.

a_1 = 3(1) - 1 = 2

Find the first term by plugging in k=1.

a_{50} = 3(50) - 1 = 149

Find the last term by plugging in k=50.

S_{50} = \frac{50}{2}(2 + 149)

n=50, first term=2, last term=149. Apply the formula.

S_{50} = 25 \cdot 151 = 3775 \checkmark

Half of 50 is 25. Multiply by the sum of first and last.

Sometimes a problem gives you the sum and asks you to find a missing piece. The formula works in reverse — treat it as an equation and solve.

Finding n when the sum is given: S_n = 270, a_1 = 3, d = 3

270 = \frac{n}{2}(3 + a_n)

Set up the formula with what you know. You still need a_n, so express it in terms of n.

a_n = 3 + (n-1)(3) = 3n

Use the sequence formula: a_n = a_1 + (n-1)d = 3 + 3(n-1) = 3n.

270 = \frac{n}{2}(3 + 3n)

Substitute the expression for a_n back in.

270 = \frac{n}{2} \cdot 3(1 + n)

Factor 3 out of the parentheses.

270 = \frac{3n(n+1)}{2}

Clean up the right side.

540 = 3n(n+1)

Multiply both sides by 2.

180 = n(n+1)

Divide both sides by 3. Now you need two consecutive integers that multiply to 180.

n = 13, \quad 13 \cdot 14 = 182 \neq 180

Try n=13. Close, but not right.

n = 12, \quad 12 \cdot 13 = 156 \neq 180

Try n=12. Also not right — so go back and check.

n^2 + n - 180 = 0

Rewrite as a quadratic and solve properly.

(n-12)(n+15) = 0 \implies n = 12 \checkmark

Factor. n=-15 is rejected because the number of terms must be positive.

Here is a Desmos graph of the partial sums of the arithmetic series $1 + 2 + 3 + \cdots$ — the Gauss problem. Each point shows $S_n$ for whole-number values of $n$ . The pattern is a parabola, which makes sense: KaTeX can only parse string typed expression, a quadratic in $n$ .

Interactive graph — scroll to zoom, drag to pan

Practice Questions

\sum_{k=1}^{6} (4k - 3)

S_{15} \text{ for the sequence } 10 + 13 + 16 + \cdots

\sum_{k=3}^{8} (5k + 2)

Regents Corner

On the Algebra II Regents, arithmetic series problems appear on both Part I and Part II. Part II problems often give you the sum and ask you to find $n$ or $d$ , which requires setting up an equation and solving it. Partial credit is available, so always write the formula you use before substituting numbers. A grader who sees KaTeX can only parse string typed expression written out before an arithmetic error can still award process credit. A blank setup earns nothing even if the arithmetic that follows is correct.

When sigma notation does not start at k=1, students still use n equal to the top index. In the sum from k=3 to k=8, there are not 8 terms — there are 6. The correct count is top minus bottom plus one: 8 - 3 + 1 = 6. Using n=8 inflates the answer and is a very common error on Part II problems involving sigma notation.

Geometric Series