Sequences, Series and Sigma Notation
Before you can tackle geometric or arithmetic series in finance and modelling, you need the pure mathematical scaffolding: what a sequence is, how partial sums work, and how sigma notation compresses a sum into elegant shorthand.
A stadium has 1 seat in row 1, 2 in row 2, 3 in row 3, and so on up to row 50. Write your gut answers — no calculating yet:
- How many seats are in row 20?
- What is the total number of seats across all 50 rows?
- Can you describe a pattern or shortcut for finding that total?
A sequence is an ordered list of terms. A series is the sum of those terms. Sigma notation is the shorthand that collapses a sum into a single expression. These three ideas are the scaffolding for every HSC series question.
Sequence: $T_1, T_2, T_3, \ldots$ — ordered, each element is a term. Partial sum: $S_n = T_1 + T_2 + \cdots + T_n$. Series: infinite sum $S = T_1 + T_2 + T_3 + \cdots$. Sigma: $\displaystyle\sum_{k=1}^{n} T_k = S_n$.
Key facts
- Definition of a sequence as an ordered list with notation $T_n$
- Difference between finite sequences (ends at $T_N$) and infinite sequences
- Definition of partial sum $S_n$ and infinite series $S$
- Components of sigma notation: $\sum$, index, lower limit, upper limit, general term
Concepts
- Why partial sums form their own sequence
- How $T_n = S_n - S_{n-1}$ connects a term to its partial sum formula
- The difference between a series that converges and one that diverges
Skills
- Write the first several terms of a sequence from an explicit formula or recurrence relation
- Compute partial sums $S_n$ directly and verify with sigma notation
- Recover $T_n$ from a given partial sum formula $S_n$, checking $T_1$ separately
- Evaluate sigma sums by expanding and adding each term
A sequence is an ordered list of objects — usually numbers — where each element occupies a specific position. The position number $n$ is always a positive integer.
Notation and term rules:
- $T_n$ notation: $T_1$ is the first term, $T_2$ the second, $T_n$ the $n$th term. The subscript is the position.
- Explicit (general term) formula: gives $T_n$ directly in terms of $n$. For $T_n = 2n$: $T_1 = 2$, $T_2 = 4$, $T_3 = 6$, $\ldots$ (every even number). For $T_n = n^2$: $T_1=1$, $T_2=4$, $T_3=9$, $\ldots$ (perfect squares).
- Recurrence (recursive) relation: defines each term using the previous one. $T_1 = 3$, $T_n = T_{n-1} + 4$ gives: $3, 7, 11, 15, \ldots$
- Finite sequence: terminates — written without ellipsis, e.g. $2, 4, 6, 8, 10$. Has $N$ terms.
- Infinite sequence: continues without end — written with ellipsis, e.g. $2, 4, 6, 8, 10, \ldots$
Sequence A: $1, 4, 9, 16, 25$ — finite (5 terms), explicit formula $T_n = n^2$.
Sequence B: $1, 2, 4, 8, 16, \ldots$ — infinite, explicit formula $T_n = 2^{n-1}$.
Sequence C: $T_1 = 5$, $T_n = T_{n-1} - 3$ — recurrence relation giving $5, 2, -1, -4, \ldots$
Which has $T_5 = 32$? Sequence B: $T_5 = 2^{5-1} = 2^4 = 16$ — no. Check Sequence B: $T_6 = 2^5 = 32$, so $T_6 = 32$ in B. Actually $T_n = 2^{n-1}$, so $T_n = 32 \Rightarrow 2^{n-1} = 32 = 2^5 \Rightarrow n = 6$.
Sequence = ordered list of terms. $T_n$ = $n$th term ($n$ is a positive integer).; Explicit formula: $T_n = f(n)$ gives the term directly. Recurrence: $T_n = g(T_{n-1})$ defines each term from the previous one.
Pause — copy the definitions: sequence = ordered list of terms; $T_n$ = $n$th term; explicit formula gives $T_n$ directly from $n$; recurrence defines $T_n$ from the previous term $T_{n-1}$ — into your book.
Quick check: A sequence is defined by $T_1 = 3$ and $T_n = T_{n-1} + 4$ for $n \geq 2$. What is $T_4$?
We just saw that a sequence is an ordered list of terms described by an explicit formula $T_n = f(n)$ or a recurrence relation. That raises a question: given the terms, how do we compute their running total — and can we recover any individual term from the total? This card answers it → the partial sum $S_n = T_1 + \cdots + T_n$ is itself a new sequence, and the identity $T_n = S_n - S_{n-1}$ recovers any term from consecutive partial sums.
The $n$th partial sum $S_n$ is the sum of the first $n$ terms of a sequence. Partial sums themselves form a new sequence — and understanding their relationship to $T_n$ is critical for HSC problems.
Computing partial sums — example with $T_n = 2n$:
- $S_1 = T_1 = 2$
- $S_2 = T_1 + T_2 = 2 + 4 = 6$
- $S_3 = T_1 + T_2 + T_3 = 2 + 4 + 6 = 12$
- $S_4 = 2 + 4 + 6 + 8 = 20$
- $S_5 = 2 + 4 + 6 + 8 + 10 = 30$
The partial sums $2, 6, 12, 20, 30, \ldots$ form their own sequence. Notice the closed-form: $S_n = n(n+1)$ in this case.
Key identity: For any $n \geq 2$, $T_n = S_n - S_{n-1}$. This makes sense: $S_n$ includes $T_n$ but $S_{n-1}$ does not, so the difference isolates $T_n$.
$S_n = T_1 + T_2 + \cdots + T_n$ — the sum of the first $n$ terms.; $T_n = S_n - S_{n-1}$ for $n \geq 2$. Always check $T_1 = S_1$ separately.
Pause — copy the partial sum definition $S_n = T_1 + T_2 + \cdots + T_n$ and the key identity $T_n = S_n - S_{n-1}$ for $n \geq 2$ (always check $T_1 = S_1$ separately) into your book.
Fill the blanks: For the sequence $T_n = 3n$, the first four terms are $T_1 = 3$, $T_2 = 6$, $T_3 = 9$, $T_4 = 12$. Therefore $S_4 = $ and $S_3 = $ . Using the identity, $T_4 = S_4 - S_3 = $ .
We just saw the partial sum $S_n$ as the cumulative total of the first $n$ terms. That raises a question: what happens when the sequence is infinite — does adding infinitely many terms always blow up, or can the total converge to a finite value? This card answers it → a finite series is just $S_N$; an infinite series is $\lim_{n\to\infty} S_n$, which may converge (like $1 + \frac{1}{2} + \frac{1}{4} + \cdots = 2$) or diverge (like $1 + 2 + 3 + \cdots$).
A series is the sum of the terms of a sequence. When a sequence is finite, the series is simply $S_N$. When the sequence is infinite, the series is a limit — it may or may not settle on a finite value.
Types of series:
- Finite series: $S_N = T_1 + T_2 + \cdots + T_N$ — a definite number. The stadium problem is a finite series: $S_{50} = 1 + 2 + \cdots + 50 = 1275$.
- Infinite series: $S = \displaystyle\lim_{n \to \infty} S_n = T_1 + T_2 + T_3 + \cdots$ — a limit. This may exist (converge) or not (diverge).
- Diverging series example: $1 + 2 + 3 + \cdots$ — partial sums grow without bound. No finite sum exists.
- Converging series example (preview): $1 + \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \cdots$ — partial sums approach 2. The infinite sum is exactly 2. (Full convergence theory is Lesson 4.)
Finite series = sum of all terms of a finite sequence = $S_N$.; Infinite series = $\lim_{n\to\infty} S_n$ — may converge (finite sum) or diverge (no finite sum).
Pause — copy the distinction: finite series = $S_N$; infinite series = $\lim_{n\to\infty} S_n$ — may converge (geometric with $|r| < 1$) or diverge ($1+2+3+\cdots$ has no finite limit) — into your book.
True or false: Every infinite series has a finite sum.
We just saw that a series is a sum of terms, finite or infinite. That raises a question: writing out $T_1 + T_2 + \cdots + T_n$ becomes cumbersome for long sums — is there a compact notation that specifies the general term, starting index, and ending index in one expression? This card answers it → sigma notation $\displaystyle\sum_{k=r}^{n} T_k$ compresses the entire sum into a single symbol.
Sigma notation $\displaystyle\sum_{k=r}^{n} T_k$ compresses a sum into a single expression. Reading and writing sigma notation fluently is essential for every HSC series topic.
Anatomy of sigma notation:
- $\Sigma$ (capital sigma): means "sum".
- Lower limit ($k = r$): the starting value of the index. Most commonly $k = 1$.
- Upper limit ($n$): the ending value of the index. Can be a specific number or $\infty$ for an infinite series.
- General term ($T_k$): the expression you substitute $k$ into for each step.
Examples with computed values:
- $\displaystyle\sum_{k=1}^{4}(2k-1) = 1 + 3 + 5 + 7 = 16$ (sum of first 4 odd numbers)
- $\displaystyle\sum_{k=3}^{7}k^2 = 9 + 16 + 25 + 36 + 49 = 135$ (sum starts at $k=3$, not $k=1$)
- $\displaystyle\sum_{k=1}^{10}(2k-1) = $ sum of first 10 odd numbers $= 1+3+5+\cdots+19 = 100$
- $\displaystyle\sum_{k=1}^{\infty}\frac{1}{2^k} = \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots = 1$ (converging infinite series)
"The sum of the first 10 odd numbers" $\to$ $\displaystyle\sum_{k=1}^{10}(2k-1)$
"The sum of the squares from 1 to 6" $\to$ $\displaystyle\sum_{k=1}^{6}k^2$
"The sum of all terms of an infinite geometric sequence with first term 1 and ratio $\tfrac{1}{3}$" $\to$ $\displaystyle\sum_{k=1}^{\infty}\left(\tfrac{1}{3}\right)^{k-1}$
$\displaystyle\sum_{k=r}^{n} T_k$: sum the general term $T_k$ for $k = r, r+1, \ldots, n$.; Infinite series: upper limit is $\infty$. May converge or diverge.
Pause — copy sigma notation $\displaystyle\sum_{k=r}^{n} T_k$ (lower limit $r$, upper limit $n$, index $k$ is a dummy variable) and the infinite form $\displaystyle\sum_{k=r}^{\infty} T_k$ into your book.
Match each sigma expression with its expanded form and value:
We just saw sigma notation as a compact way to write any sum. That raises a question: the HSC often gives you a formula for $S_n$ and asks you to find $T_n$ — how do you extract the general term from a sum formula, and what can go wrong at $n = 1$? This card answers it → use $T_n = S_n - S_{n-1}$ for $n \geq 2$, then verify $T_1 = S_1$; if the check fails, state $T_1$ as a separate piecewise value.
A common HSC question gives you a formula for $S_n$ (the partial sum) and asks you to find $T_n$ (the general term). The technique is $T_n = S_n - S_{n-1}$, with a separate check for $T_1$.
Worked example: $S_n = n^2 + 3n$
- Step 1 — find $T_n$ for $n \geq 2$: $$T_n = S_n - S_{n-1} = (n^2 + 3n) - \bigl((n-1)^2 + 3(n-1)\bigr)$$ $$= n^2+3n - (n^2-2n+1+3n-3) = n^2+3n - n^2 - n + 2 = 2n+2$$
- Step 2 — check $T_1$ using $S_1$: $S_1 = 1^2 + 3(1) = 4$. Does the formula give $T_1 = 2(1)+2 = 4$? Yes ✓. So $T_n = 2n+2$ works for all $n \geq 1$.
- Step 3 — find specific terms: $T_5 = 2(5)+2 = 12$. Verify: $S_5 = 25+15 = 40$ and $S_4 = 16+12 = 28$. $S_5 - S_4 = 12$ ✓.
Given $S_n$: use $T_n = S_n - S_{n-1}$ for $n \geq 2$ to find the general term.; Always check: does $T_1 = S_1$? If yes, formula holds for all $n$. If no, state $T_1$ separately.
Pause — copy the technique: $T_n = S_n - S_{n-1}$ for $n \geq 2$; always verify $T_1 = S_1$; if the check fails, state $T_1$ separately as a piecewise formula — into your book.
Top 3 list: $S_n = 3n^2$. List THREE key working steps you would use to find $T_n$ and verify your answer.
Worked examples · 3 problems
A sequence is defined by $T_n = n^2 - n + 1$. (a) Write the first five terms. (b) Find $S_5$. (c) Express $S_5$ using sigma notation.
$T_1 = 1-1+1 = 1$
$T_2 = 4-2+1 = 3$
$T_3 = 9-3+1 = 7$
$T_4 = 16-4+1 = 13$
$T_5 = 25-5+1 = 21$
$S_5 = T_1 + T_2 + T_3 + T_4 + T_5$
$= 1 + 3 + 7 + 13 + 21 = \mathbf{45}$
$S_5 = \displaystyle\sum_{k=1}^{5}(k^2 - k + 1) = 45$
Evaluate $\displaystyle\sum_{k=2}^{5}(3k - 1)$ by expanding each term and summing.
(four terms total)
$k=2$: $3(2)-1 = 5$
$k=3$: $3(3)-1 = 8$
$k=4$: $3(4)-1 = 11$
$k=5$: $3(5)-1 = 14$
$5 + 8 + 11 + 14 = \mathbf{38}$
The sum of the first $n$ terms of a sequence is $S_n = 4n^2 - n$. (a) Find a formula for the general term $T_n$ for $n \geq 2$. (b) Verify that the formula holds for $T_1$. (c) Find $T_7$.
$S_n = 4n^2 - n$
$S_{n-1} = 4(n-1)^2 - (n-1) = 4(n^2-2n+1)-(n-1)$
$= 4n^2 - 8n + 4 - n + 1 = 4n^2 - 9n + 5$
$T_n = (4n^2-n) - (4n^2-9n+5) = 8n - 5$
$S_1 = 4(1)^2 - 1 = 3$
Formula: $T_1 = 8(1) - 5 = 3$ ✓
The formula $T_n = 8n-5$ holds for all $n \geq 1$.
$T_7 = 8(7) - 5 = 56 - 5 = \mathbf{51}$
Verify: $S_7 = 4(49)-7=196-7=189$
$S_6 = 4(36)-6=144-6=138$
$T_7 = 189-138 = 51$ ✓
The stadium has $S_{50} = 1 + 2 + 3 + \cdots + 50 = 1275$ seats total. In sigma notation: $\displaystyle\sum_{k=1}^{50} k = 1275$. The "shortcut" is the arithmetic series formula $S_n = \frac{n}{2}(T_1 + T_n)$ — which you will prove rigorously in Lesson 3. Now revisit your initial intuitions:
Q1. The sequence $5, 8, 11, 14, \ldots$ has general term $T_n = ?$
Q2. $\displaystyle\sum_{k=1}^{4}(2k-1) = ?$
Q3. If $S_n = 3n^2$, then $T_3 = ?$
Q4. A sequence has $T_1 = 2$ and $T_n = T_{n-1} + 3$ for $n \geq 2$. What is $T_6$?
Q5. Which of the following is an infinite series?
SA 1. A sequence is defined by $T_1 = 5$ and $T_n = T_{n-1} - 3$ for $n \geq 2$. (a) Write the first five terms of the sequence. (b) Find $S_5$. (3 marks)
SA 2. Evaluate $\displaystyle\sum_{k=2}^{6}(k^2 - 1)$ by expanding and adding each term. Express your answer as a single integer. (3 marks)
SA 3. The partial sum of a sequence is given by $S_n = 2n^2 + n$. (a) Find a formula for $T_n$ for $n \geq 2$. (b) Verify the formula holds for $T_1$. (c) Find the value of $n$ for which $T_n = 35$. (4 marks)
📖 Comprehensive answers (click to reveal)
SA 1 (3 marks): (a) $T_1 = 5$, $T_2 = 5-3 = 2$, $T_3 = 2-3 = -1$, $T_4 = -1-3 = -4$, $T_5 = -4-3 = -7$ [1 mark for all 5 correct]. (b) $S_5 = 5 + 2 + (-1) + (-4) + (-7) = 5 + 2 - 1 - 4 - 7 = -5$ [1 mark method, 1 mark answer].
SA 2 (3 marks): $k=2$: $4-1=3$; $k=3$: $9-1=8$; $k=4$: $16-1=15$; $k=5$: $25-1=24$; $k=6$: $36-1=35$ [1 mark for correct expansion]. Sum $= 3+8+15+24+35 = 85$ [1 mark method, 1 mark answer].
SA 3 (4 marks): (a) $T_n = S_n - S_{n-1} = (2n^2+n) - (2(n-1)^2+(n-1)) = (2n^2+n) - (2n^2-4n+2+n-1) = (2n^2+n) - (2n^2-3n+1) = 4n-1$ [2 marks]. (b) $S_1 = 2+1 = 3$. Formula: $T_1 = 4(1)-1 = 3$ ✓ [1 mark]. (c) $4n-1 = 35 \Rightarrow 4n = 36 \Rightarrow n = 9$ [1 mark].
Five timed questions on sequences, series and sigma notation. Gold tier: 90% + speed.
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