Sequences, partial sums and sigma notation: reading and writing sums with the Greek letter sigma, the link a_n = S_n - S_(n-1), and linking arithmetic and geometric sums to sigma form (new focus area in the 2024 Mathematics Advanced syllabus)

Syllabus dot point

“Use sequence notation $a_n$ , partial sums $S_n$ and summation (sigma) notation $\sum_{k=i}^{j} a_k$ to describe, evaluate and manipulate the sum of terms of a sequence, including infinite series that converge”

HSCMaths AdvancedYear 12: Sequences and Series12 min readUpdated 7 Oct 2026

Quick answer

Sigma notation $\sum_{k=i}^{j} a_k$ adds the terms $a_i$ to $a_j$ ( $j - i + 1$ of them). The partial sum is $S_n$ , a term is $a_n = S_n - S_{n-1}$ (with $a_1 = S_1$ ), arithmetic series give $\frac{n}{2}(2a + (n-1)d)$ , geometric series give $\frac{a(1 - r^n)}{1 - r}$ and converge to $\frac{a}{1-r}$ when $|r| < 1$ . Sequences and series is its own Year 12 focus area in the 2024 syllabus.

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What this dot point is asking
The answer
Exam-style questions
Practice questions

What this dot point is asking

In the Mathematics Advanced 11-12 Syllabus (2024), Sequences and series is a Year 12 focus area of its own, no longer a sub-topic hidden inside Financial mathematics. The 2017 course taught arithmetic and geometric sequences mainly through finance. The new content begins one step earlier, with the general ideas: what a sequence is, what a partial sum is, and how to write a sum using summation (sigma) notation. NESA describes a series as the sum of the terms of a sequence and expects you to use $\sum$ to express sums of terms from position $i$ to position $j$ .

So the job is to be fluent in three skills: read sigma notation (turn it into a list of terms), write it (from a pattern), and evaluate it (using the arithmetic and geometric formulas you already know).

Note

A sequence is just a list of numbers in order. A sum of the first few of them is called a partial sum. Sigma notation is shorthand for "add up this pattern", the way a recipe says "repeat for each egg". The letter underneath is a counter that starts at the bottom number, steps by one, and stops at the top number. Whatever is written to the right of the sigma tells you what to do with the counter each time.

The answer

Key fact

A sequence is an ordered list $a_1, a_2, a_3, \dots$ with $n$ th term $a_n$ . A finite sequence stops; an infinite one does not.
The partial sum is $S_n = a_1 + a_2 + \dots + a_n$ .
Sigma notation: $\displaystyle\sum_{k=i}^{j} a_k = a_i + a_{i+1} + \dots + a_j$ , a total of $j - i + 1$ terms.
A term is a difference of partial sums: $a_n = S_n - S_{n-1}$ for $n \geq 2$ , and $a_1 = S_1$ .

Reading sigma notation

Three parts matter: the lower limit (where $k$ starts), the upper limit (where it stops) and the general term (what you evaluate each time). For example

\sum_{k=2}^{5} (k^2 - 1) = (4 - 1) + (9 - 1) + (16 - 1) + (25 - 1) = 3 + 8 + 15 + 24 = 50.

The letter $k$ is a dummy variable. $\sum_{k=2}^{5}(k^2-1)$ and $\sum_{m=2}^{5}(m^2-1)$ mean exactly the same thing, and neither contains a variable once evaluated.

Writing sigma notation from a pattern

Write out the first few terms and find a rule in terms of the position number.
Decide where to start the counter. Starting at $1$ is usual, but any start works if the rule is adjusted.
Count the terms to find the upper limit.

For $5 + 8 + 11 + 14 + 17$ the terms rise by $3$ , so the rule is $3k + 2$ with $k = 1$ to $5$ . The same sum is $\sum_{k=0}^{4} (3k + 5)$ , which shows that the limits and the rule move together.

Properties of sigma

These follow directly from adding term by term, so you can use them without proof:

\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} c = nc.

You can also split a sum at any point: $\sum_{k=1}^{n} a_k = \sum_{k=1}^{m} a_k + \sum_{k=m+1}^{n} a_k$ . Rearranged, this gives a quick way to evaluate a sum that does not start at $1$ : $\sum_{k=m+1}^{n} a_k = S_n - S_m$ .

Sigma form of arithmetic and geometric series

An arithmetic sequence has $a_k = a + (k-1)d$ , so

\sum_{k=1}^{n} \left(a + (k-1)d\right) = \frac{n}{2}\left(2a + (n-1)d\right) = \frac{n}{2}(a + a_n).

A geometric sequence has $a_k = ar^{k-1}$ , so for $r \neq 1$

\sum_{k=1}^{n} ar^{k-1} = \frac{a(1 - r^n)}{1 - r}.

NESA expects you to see where the geometric formula comes from. Multiply $x^{n-1} + x^{n-2} + \dots + x + 1$ by $(x - 1)$ : every middle term cancels and the product is $x^n - 1$ . Setting $x = r$ and scaling by $a$ gives the formula above.

When $|r| < 1$ the term $r^n$ tends to $0$ , so the partial sums approach a limiting sum:

\sum_{k=1}^{\infty} ar^{k-1} = \frac{a}{1 - r}.

Finding a term from partial sums

If you are given $S_n$ rather than $a_n$ , subtract:

a_n = S_n - S_{n-1}.

The reason is that $S_n$ contains all the terms up to $a_n$ and $S_{n-1}$ contains all but the last. This is the standard route when a question gives "the sum of the first $n$ terms is ..." and asks for a particular term or to prove that the sequence is arithmetic or geometric.

Worked examples

Evaluate a sum that starts above 1

Evaluate $\displaystyle\sum_{k=5}^{12} (3k + 1)$ .

Find the partial sum formula

The terms $3k + 1$ for $k = 1, \dots, n$ form an arithmetic sequence with first term $4$ and difference $3$ , so $S_n = \frac{n}{2}(8 + 3(n-1)) = \frac{n(3n+5)}{2}$ .

Subtract

$S_{12} = \frac{12 \times 41}{2} = 246$ and $S_4 = \frac{4 \times 17}{2} = 34$ , so the required sum is $246 - 34 = 212$ .

Marker's note: the sum from $k = 5$ removes the first four terms, so it is $S_{12} - S_4$ .

Prove a sequence is geometric from its partial sums

The partial sum of a sequence is $S_n = 3^n - 1$ . Find $a_n$ and state what type of sequence it is.

Take the difference

For $n \geq 2$ , $a_n = (3^n - 1) - (3^{n-1} - 1) = 3^n - 3^{n-1} = 3^{n-1}(3 - 1) = 2 \cdot 3^{n-1}$ . Also $a_1 = S_1 = 2$ , which fits the same rule.

Identify

The ratio of consecutive terms is $\frac{a_{n+1}}{a_n} = 3$ , constant, so the sequence is geometric with $a = 2$ and $r = 3$ .

Marker's note: showing the constant ratio is the proof; do not just state the type.

Common traps

Miscounting the terms: $\sum_{k=3}^{9}$ has $9 - 3 + 1 = 7$ terms, not $6$ and not $9$ .
Using the wrong partial sum to start a block: To sum from $k = m$ to $n$ you need $S_n - S_{m-1}$ , not $S_n - S_m$ .
Forgetting $a_1 = S_1$: The rule $a_n = S_n - S_{n-1}$ needs $n \geq 2$ . Check the first term separately.
Summing an infinite series with $|r| \geq 1$: The formula $\frac{a}{1-r}$ only applies when $|r| < 1$ . Always state this condition.
Treating the dummy letter as a variable: The answer to a fully evaluated sigma has no $k$ in it.

Exam technique

Write out the first two or three terms of any unfamiliar sigma before choosing a method: that tells you whether the series is arithmetic, geometric or neither. If you are asked to "show that" a sum equals a formula, split it using the properties of sigma and quote $\sum k = \frac{n(n+1)}{2}$ or the standard arithmetic or geometric result. For the limiting sum, state $|r| < 1$ before you apply the formula.

Exam-style questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

HSC-style3 marks

The

n

th term of a sequence is

a_n = 5n - 3

. Find

\displaystyle\sum_{k=1}^{20} a_k

Show worked answer →

The terms $2, 7, 12, \dots$ form an arithmetic sequence with first term $a = 2$ and common difference $d = 5$ . Using $S_n = \frac{n}{2}(2a + (n-1)d)$ with $n = 20$ :

S_{20} = 10\left(4 + 19 \times 5\right) = 10 \times 99 = 990.

Markers give one mark for identifying $a$ and $d$ , one for a correct formula and one for the value. A common slip is using $n = 19$ because the upper limit is mistaken for a count starting at zero.

HSC-style3 marks

Express

\displaystyle\sum_{k=4}^{10} (2k - 1)

as a difference of two sums starting at

k = 1

, and evaluate it.

Show worked answer →

Remove the first three terms:

\sum_{k=4}^{10} (2k-1) = \sum_{k=1}^{10} (2k-1) - \sum_{k=1}^{3} (2k-1) = 10^2 - 3^2 = 100 - 9 = 91,

using the fact that the sum of the first $n$ odd numbers is $n^2$ . Checking directly: $7 + 9 + 11 + 13 + 15 + 17 + 19 = 91$ . Markers reward the split into two sums (the number of terms removed must be $3$ , not $4$ ) and a correct evaluation.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks

Evaluate

\displaystyle\sum_{k=1}^{5} (3k - 2)

Show worked solution →

Substitute $k = 1, 2, 3, 4, 5$ in turn and add.

\sum_{k=1}^{5} (3k - 2) = 1 + 4 + 7 + 10 + 13 = 35.

Marker's note: one mark for the correct five terms, one for the total. Listing the terms is safer than jumping to a formula.

foundation2 marks

Write

4 + 9 + 16 + 25 + 36

in sigma notation.

Show worked solution →

Spot the pattern. The terms are $2^2, 3^2, 4^2, 5^2, 6^2$ , so the general term is $k^2$ and $k$ runs from $2$ to $6$ .

4 + 9 + 16 + 25 + 36 = \sum_{k=2}^{6} k^2.

Marker's note: one mark for the general term, one for the correct limits. An equally correct answer is $\sum_{k=1}^{5} (k+1)^2$ .

core3 marks

Evaluate

\displaystyle\sum_{k=1}^{8} 3 \cdot 2^{k-1}

Show worked solution →

Recognise a geometric series. The first term is $a = 3 \cdot 2^0 = 3$ and the common ratio is $r = 2$ , with $n = 8$ terms.

Use $S_n = \dfrac{a(r^n - 1)}{r - 1}$ (the form that suits $r > 1$ ):

S_8 = \frac{3(2^8 - 1)}{2 - 1} = 3 \times 255 = 765.

Marker's note: one mark for identifying $a$ , $r$ and $n$ , one for the formula, one for the answer.

core3 marks

The sum of the first

n

terms of a sequence is

S_n = n^2 + 2n

. (a) Find a formula for the

n

th term

a_n

. (b) Check your formula by showing that

a_7 + a_8 + a_9 + a_{10} = S_{10} - S_6

Show worked solution →

(a) A term is the difference of two partial sums. For $n \geq 2$ , $a_n = S_n - S_{n-1}$ :

a_n = \left(n^2 + 2n\right) - \left((n-1)^2 + 2(n-1)\right) = n^2 + 2n - n^2 + 2n - 1 - 2n + 2 = 2n + 1.

Also $a_1 = S_1 = 3$ , which agrees with $2(1) + 1 = 3$ , so $a_n = 2n + 1$ for all $n \geq 1$ .

(b) Compare both sides. $S_{10} - S_6 = (100 + 20) - (36 + 12) = 120 - 48 = 72$ . Directly, $a_7 + a_8 + a_9 + a_{10} = 15 + 17 + 19 + 21 = 72$ . The two agree.

Marker's note: one mark for $a_n = S_n - S_{n-1}$ , one for the simplified $2n + 1$ , one for the check. Always test $n = 1$ separately.

exam4 marks

(a) Show that

\displaystyle\sum_{k=1}^{n} (4k - 1) = 2n^2 + n

. (b) Hence find the value of

n

for which the sum equals

300

Show worked solution →

(a) Split the sum using the properties of sigma.

\sum_{k=1}^{n} (4k - 1) = 4\sum_{k=1}^{n} k - \sum_{k=1}^{n} 1 = 4 \cdot \frac{n(n+1)}{2} - n = 2n^2 + 2n - n = 2n^2 + n.

(Equivalently, the terms $3, 7, 11, \dots$ form an arithmetic sequence with $a = 3$ and $d = 4$ , so $S_n = \frac{n}{2}\left(2 \cdot 3 + (n-1) \cdot 4\right) = 2n^2 + n$ .)

(b) Solve $2n^2 + n = 300$ . Then $2n^2 + n - 300 = 0$ , so $(2n + 25)(n - 12) = 0$ . Since $n$ is a positive whole number, $n = 12$ .

Check: $2(144) + 12 = 300$ .

Marker's note: one mark for splitting the sum, one for $\sum k = \frac{n(n+1)}{2}$ , one for the equation, one for rejecting $n = -\frac{25}{2}$ .

exam4 marks

(a) Find

\displaystyle\sum_{k=1}^{\infty} 6\left(\tfrac{1}{3}\right)^{k}

. (b) Find the smallest

n

for which the partial sum

S_n

of this series exceeds

2.99

Show worked solution →

(a) A geometric series. The first term is $a = 6 \cdot \frac{1}{3} = 2$ and the ratio is $r = \frac{1}{3}$ . Since $|r| < 1$ the series converges:

S_\infty = \frac{a}{1 - r} = \frac{2}{1 - \frac{1}{3}} = \frac{2}{\frac{2}{3}} = 3.

(b) Write the partial sum. $S_n = \dfrac{2\left(1 - \left(\frac{1}{3}\right)^n\right)}{1 - \frac{1}{3}} = 3\left(1 - \left(\tfrac{1}{3}\right)^n\right)$ .

Require $3\left(1 - 3^{-n}\right) > 2.99$ , so $3^{-n} < \frac{0.01}{3} = \frac{1}{300}$ , that is $3^n > 300$ .

Test: $3^5 = 243 < 300$ but $3^6 = 729 > 300$ . So the smallest $n$ is $6$ (indeed $S_5 = \frac{242}{81} \approx 2.988$ and $S_6 = \frac{728}{243} \approx 2.996$ ).

Marker's note: one mark for $a$ and $r$ , one for $S_\infty = 3$ , one for the inequality in terms of $3^n$ , one for testing consecutive powers.

Practise this

Sources & how we know this

Mathematics Advanced 11-12 Syllabus (2024): Year 12 Sequences and series — NESA (2024)
Mathematics Advanced 11-12 Syllabus (2024): course overview and implementation — NESA (2024)
Mathematics Advanced Stage 6 Syllabus (2017) — NESA
Mathematics Advanced HSC exam papers — NESA