Series

Lessons

Senior 6 · Mathematics

Series

4 lessons

Lesson 1 of 4

Finite series and sigma notation

25 min

A long sum, written short

How would you add $1^2 + 2^2 + 3^2 + \cdots + 100^2$ ? Writing out a hundred terms is slow, and adding them is slower. This lesson gives a compact notation for such sums and formulas that evaluate them in one line. In S5 (Sequences) you studied arithmetic and geometric sequences and the sums of their first $n$ terms. Here those sums become series, and the unit goes on to add infinitely many terms, which is one way a calculator can work out $e^{x}$ , $\sin x$ and $\ln x$ .

Definition

Series: the sum of the terms of a sequence. A finite series has a last term; an infinite series does not.

Sigma notation: for a sequence $u_1, u_2, u_3, \ldots$ ,

\sum_{r=m}^{n} u_r = u_m + u_{m+1} + \cdots + u_n ,

read "the sum of $u_{r}$ from $r = m$ to $r = n$ ". The letter $r$ is the index (any letter will do), and $m$ and $n$ are the lower and upper limits. The sum has $n - m + 1$ terms.

Example: $\sum_{r=1}^{4} r^2 = 1 + 4 + 9 + 16 = 30$ , and $3 + 7 + 11 + \cdots + 43 = \sum_{r=1}^{11}(4r - 1)$ .

Important

Rules of sigma notation (they follow from regrouping a finite sum):

$\sum_{r=1}^{n}(a u_r + b v_r) = a\sum_{r=1}^{n} u_r + b\sum_{r=1}^{n} v_r$ for constants $a, b$ .
$\sum_{r=1}^{n} c = nc$ for a constant $c$ : the constant is added $n$ times.
$\sum_{r=m}^{n} u_r = \sum_{r=1}^{n} u_r - \sum_{r=1}^{m-1} u_r$ .

Arithmetic and geometric series

Arithmetic series, first term $a$ , common difference $d$ . Write $S_{n}$ forwards and backwards and add: each of the $n$ pairs of terms adds to $a + \ell$ , where $\ell = a + (n - 1)d$ is the last term. So $2S_n = n(a + \ell)$ and

S_n = \frac{n}{2}(a + \ell) = \frac{n}{2}\left[2a + (n - 1)d\right] .

Geometric series, first term $a$ , common ratio $r \ne 1$ . Subtract $rS_n = ar + ar^2 + \cdots + ar^n$ from $S_n = a + ar + \cdots + ar^{n-1}$ : all but two terms cancel, so $(1 - r) S_{n} = a - a r^{n}$ and

S_n = \frac{a(1 - r^n)}{1 - r} .

Worked example 1: evaluating sums

Evaluate (a) $\sum_{r=1}^{20}(3r + 2)$ (b) $\sum_{r=0}^{9} 5 \times 2^r$ , and (c) write $2 + 6 + 18 + \cdots + 2 \times 3^9$ in sigma notation and evaluate it.

Solution

Arithmetic, 20 terms, first 5, last 62: $\frac{20}{2}(5 + 62) = 670$ . Or by the rules: $3\sum r + \sum 2 = 3 \times 210 + 40 = 670$ .
Geometric, 10 terms, $a = 5$ , $r = 2$ : $\frac{5(2^{10} - 1)}{2 - 1} = 5115$ .
$\sum_{r=0}^{9} 2 \times 3^r = \frac{2(3^{10} - 1)}{3 - 1} = 3^{10} - 1 = 59\,048$ .

The standard sums

Three sums are used so often that they are learned as formulas:

\sum_{r=1}^{n} r = \frac{n(n + 1)}{2}, \qquad \sum_{r=1}^{n} r^2 = \frac{n(n + 1)(2n + 1)}{6},

\sum_{r=1}^{n} r^3 = \left[\frac{n(n + 1)}{2}\right]^2 .

The first is an arithmetic series, and the picture below shows it without algebra.

Two copies of the staircase $1 + 2 + 3 + 4 + 5$ fill a $5 \times 6$ rectangle, so $1 + 2 + \cdots + 5 = \frac{5 \times 6}{2} = 15$ . The same picture works for any $n$ .

Important

Proof by induction that $\sum_{r=1}^{n} r^2 = \frac{n(n + 1)(2n + 1)}{6}$ for every integer $n \ge 1$ .

Base case $n = 1$ : the left side is $1^{2} = 1$ ; the right side is $\frac{1 \times 2 \times 3}{6} = 1$ .
Inductive step: assume $\sum_{r=1}^{k} r^2 = \frac{k(k + 1)(2k + 1)}{6}$ . Then $\begin{aligned} \sum_{r=1}^{k+1} r^2 &= \frac{k(k + 1)(2k + 1)}{6} + (k + 1)^2 \\ &= \frac{(k + 1)\left[k(2k + 1) + 6(k + 1)\right]}{6} \\ &= \frac{(k + 1)(2k^2 + 7k + 6)}{6} \\ &= \frac{(k + 1)(k + 2)(2k + 3)}{6}, \end{aligned}$ which is the formula with $n = k + 1$ .
By the principle of mathematical induction, the formula holds for all $n \ge 1$ . ∎

Worked example 2: using the standard sums

(a) Find $\sum_{r=1}^{n} r(r + 2)$ in factorised form. (b) Evaluate $\sum_{r=11}^{20} r^2$ .

Solution

$\sum r^2 + 2\sum r = \frac{n(n + 1)(2n + 1)}{6} + n(n + 1) = \frac{n(n + 1)\left[(2n + 1) + 6\right]}{6} = \frac{n(n + 1)(2n + 7)}{6}$ . Check with $n = 2$ : $1 \times 3 + 2 \times 4 = 11$ and $\frac{2 \times 3 \times 11}{6} = 11$ ✓.
$\sum_{r=1}^{20} r^2 - \sum_{r=1}^{10} r^2 = \frac{20 \times 21 \times 41}{6} - \frac{10 \times 11 \times 21}{6} = 2870 - 385 = 2485$ .

Definition

Method of differences (telescoping sum): if each term can be written as a difference $u_{r} = f (r) - f (r + 1)$ , then almost every term cancels:

\begin{aligned} \sum_{r=1}^{n} u_r &= \left[f(1) - f(2)\right] + \left[f(2) - f(3)\right] \\ &\quad + \cdots + \left[f(n) - f(n + 1)\right] \\ &= f(1) - f(n + 1) . \end{aligned}

Example: partial fractions (S4) give $\frac{1}{r(r + 1)} = \frac{1}{r} - \frac{1}{r + 1}$ , so $\sum_{r=1}^{n}\frac{1}{r(r + 1)} = 1 - \frac{1}{n + 1} = \frac{n}{n + 1}$ .

Warning

Count the terms. $\sum_{r=3}^{10} u_r$ has $10 - 3 + 1 = 8$ terms, not 7; so $\sum_{r=3}^{10} 5 = 40$ .
Sums do not multiply. $\sum r^2 \ne \left(\sum r\right)^2$ in general: for $n = 2$ , $1 + 4 = 5$ but $(1 + 2)^{2} = 9$ .

Tip

After finding a formula for $\sum_{r=1}^{n} u_r$ , test it with $n = 1$ and $n = 2$ by adding the terms directly. A formula that fails either test is wrong, and the test takes less than a minute.

Eddie Woo: Induction with Sigma Notation, part 1 of 4, Basic example. Watch how the sum up to k + 1 is split into the sum up to k plus one extra term, the key move in the proof above.

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Practice

Questions 5 to 8 are exam length.

Write in sigma notation: (a) $3 + 7 + 11 + \cdots + 43$ (b) $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{1024}$ .
Evaluate $\sum_{r=1}^{30}(2r - 5)$ .
Evaluate $\sum_{r=1}^{8} 3 \times 2^{r-1}$ .
Evaluate $\sum_{r=1}^{10} r^2$ and $\sum_{r=1}^{10} r^3$ .
Show that $\sum_{r=1}^{n}(6r^2 - 2r) = 2n^2(n + 1)$ .
Use the method of differences to show that $\sum_{r=1}^{n}\frac{1}{(2r - 1)(2r + 1)} = \frac{n}{2n + 1}$ .
Prove by induction that $\sum_{r=1}^{n} r^3 = \frac{n^2(n + 1)^2}{4}$ .
A hall has 12 seats in its first row, and each row has 2 more seats than the row in front (illustrative figures). (a) How many seats are in 20 rows? (b) What is the least number of rows that gives at least 1000 seats?

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(a) $\sum_{r=1}^{11}(4r - 1)$ (b) $\sum_{r=1}^{10}\left(\frac{1}{2}\right)^r$ .
$2 \times 465 - 30 \times 5 = 780$ .
$\frac{3(2^8 - 1)}{2 - 1} = 765$ .
$\frac{10 \times 11 \times 21}{6} = 385$ and $\left(\frac{10 \times 11}{2}\right)^2 = 3025$ .
$6 \times \frac{n(n + 1)(2n + 1)}{6} - 2 \times \frac{n(n + 1)}{2} = n(n + 1)(2n + 1 - 1) = 2n^2(n + 1)$ .
$\frac{1}{(2r - 1)(2r + 1)} = \frac{1}{2}\left(\frac{1}{2r - 1} - \frac{1}{2r + 1}\right)$ ; the sum telescopes to $\frac{1}{2}\left(1 - \frac{1}{2n + 1}\right) = \frac{n}{2n + 1}$ .
Base case: $1 = \frac{1 \times 4}{4}$ . Step: $\frac{k^2(k + 1)^2}{4} + (k + 1)^3 = \frac{(k + 1)^2(k^2 + 4k + 4)}{4} = \frac{(k + 1)^2(k + 2)^2}{4}$ .
(a) $\frac{20}{2}(24 + 19 \times 2) = 620$ . (b) $\frac{n}{2}\left[24 + 2(n - 1)\right] = n(n + 11) \ge 1000$ gives $n \ge 26.6$ , so 27 rows (26 rows give 962 seats, 27 rows give 1026).

Sources and credits

Written by Ganzaa: the explanations, the proofs, the diagram, the worked examples, the practice items and the quiz questions. The unit follows the topic list of REB's S6 Core Mathematics (2026–2027); no REB text is reproduced.
Video: Induction with Sigma Notation (1 of 4: Basic example), Eddie Woo, embedded from YouTube.

1A long sum, written short

2Arithmetic and geometric series

3The standard sums

A long sum, written short

Arithmetic and geometric series

The standard sums