Sequences and Series

A sequence is an ordered list of numbers. A series is the sum of the terms of a sequence.

Adjust the parameters in the graph above to explore the relationships between variables.

Notation

$u_n$ or $a_n$ : the $n$ -th term of a sequence
$u_1$ : the first term
$d$ : common difference (arithmetic)
$r$ : common ratio (geometric)
$n$ : number of terms
$S_n$ : sum of the first $n$ terms

Arithmetic Sequences

An arithmetic sequence has a constant difference between consecutive terms:

U_\{n+1\} - u_n = d

General Term

U_n = u_1 + (n-1)d

Sum of the First $n$ Terms

S_n = \frac\{n\}\{2\}(u_1 + u_n) = \frac\{n\}\{2\}[2u_1 + (n-1)d]

The 5th term of an arithmetic sequence is $17$ and the 12th term is $38$ . Find the first term and the common difference.

U_5 = u_1 + 4d = 17 \quad \mathrm\{(1)\}

U_\{12\} = u_1 + 11d = 38 \quad \mathrm\{(2)\}

(2) $-$ (1): $7d = 21 \implies d = 3$ .

From (1): $u_1 + 12 = 17 \implies u_1 = 5$ .

Geometric Sequences

A geometric sequence has a constant ratio between consecutive terms:

\frac\{u_\{n+1\}\}\{u_n\} = r

General Term

U_n = u_1 r^\{n-1\}

Sum of the First $n$ Terms

S_n = \frac\{u_1(r^n - 1)\}\{r - 1\} = \frac\{u_1(1 - r^n)\}\{1 - r\} \quad (r \neq 1)

When $r = 1$ : $S_n = nu_1$ .

Find the 8th term and the sum of the first 8 terms of $2, 6, 18, 54, \ldots$ .

$u_1 = 2$ , $r = 3$ .

U_8 = 2 \times 3^7 = 2 \times 2187 = 4374

S_8 = \frac\{2(3^8 - 1)\}\{3 - 1\} = \frac\{2(6561 - 1)\}\{2\} = 6560

Geometric Mean

The geometric mean of $a$ and $b$ is $\sqrt{ab}$ (for positive $a, b$ ).

If $a, b, c$ are consecutive terms of a geometric sequence, then $b^2 = ac$ .

Convergence of Geometric Series

Sum to Infinity

If $|r| \lt 1$ The geometric series converges and the sum to infinity is:

S_\{\infty\} = \frac\{u_1\}\{1 - r\}

If $|r| \ge 1$ The series diverges (the sum to infinity does not exist).

Express $0.\dot{2}\dot{7}$ (i.e., $0.272727\ldots$ ) as a fraction.

0.\dot\{2\}\dot\{7\} = \frac\{27\}\{100\} + \frac\{27\}\{10000\} + \cdots

$u_1 = \dfrac{27}{100}$ , $r = \dfrac{1}{100}$ .

S_\{\infty\} = \frac\{\frac\{27\}\{100\}\}\{1 - \frac\{1\}\{100\}\} = \frac\{\frac\{27\}\{100\}\}\{\frac\{99\}\{100\}\} = \frac\{27\}\{99\} = \frac\{3\}\{11\}

Conditions for Convergence

Condition	Behaviour
$\\| r \\| \lt 1$	Converges to $\dfrac{u_1}{1-r}$
$r = 1$	Diverges (grows linearly)
$r = -1$	Oscillates, does not converge
$\\| r \\| \gt 1$	Diverges (grows exponentially)

Sigma Notation

Definition

\sum_\{i=1\}^\{n\} u_i = u_1 + u_2 + u_3 + \cdots + u_n

Properties

$\displaystyle\sum_{i=1}^{n} k = kn$
$\displaystyle\sum_{i=1}^{n} k u_i = k\sum_{i=1}^{n} u_i$
$\displaystyle\sum_{i=1}^{n} (u_i \pm v_i) = \sum_{i=1}^{n} u_i \pm \sum_{i=1}^{n} v_i$

Useful Sigma Summations

\sum_\{i=1\}^\{n\} i = \frac\{n(n+1)\}\{2\}

\sum_\{i=1\}^\{n\} i^2 = \frac\{n(n+1)(2n+1)\}\{6\}

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

Evaluate $\displaystyle\sum_{r=1}^{n} r(r+1)$ .

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

= \frac\{n(n+1)\}\{6\}\big[(2n+1) + 3\big] = \frac\{n(n+1)(2n+4)\}\{6\} = \frac\{n(n+1)(n+2)\}\{3\}

Applications

Compound Interest

If a principal $P$ is invested at a rate of $r$ per period, compounded for $n$ periods:

A = P(1 + r)^n

Population Growth

If a population $P_0$ grows at a rate of $r\%$ per year:

P_n = P_0\left(1 + \frac\{r\}\{100\}\right)^n

Depreciation

For depreciation at rate $r\%$ per year:

V_n = V_0\left(1 - \frac\{r\}\{100\}\right)^n

A car bought for USD 25000 depreciates at 15% per year. Find its value after 5 years.

V_5 = 25000(0.85)^5 \approx 25000 \times 0.4437 = 11092.50

The car is worth approximately USD 11092.50 after 5 years.

Loan Repayments

For a loan of $L$ with monthly repayment $R$ at monthly interest rate $i$ over $n$ months:

L = \frac\{R[1 - (1+i)^\{-n\}]\}\{i\}

The Binomial Theorem

Expansion of $(a + b)^n$

For positive integer $n$ :

(a + b)^n = \sum_\{k=0\}^\{n\} \binom\{n\}\{k\} a^\{n-k\} b^k

Where the binomial coefficient is:

\binom\{n\}\{k\} = \frac\{n!\}\{k!(n-k)!\} = ^nC_k

Special Cases

(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots

Pascal’s Triangle

Each entry is the sum of the two entries above it:

\begin`\{array}, \{cccccccc}` & & & 1 & & & \\ & & 1 & & 1 & & \\ & 1 & & 2 & & 1 & \\ 1 & & 3 & & 3 & & 1 \\ & 1 & & 4 & & 6 & & 4 & & 1 \end`\{array}`

Row $n$ (starting from row 0) gives the coefficients of $(a + b)^n$ .

Expand $(2x - 3)^4$ .

Using the binomial theorem:

(2x - 3)^4 = \sum_\{k=0\}^\{4\}\binom\{4\}\{k\}(2x)^\{4-k\}(-3)^k

= 1 \cdot (2x)^4 \cdot 1 + 4 \cdot (2x)^3 \cdot (-3) + 6 \cdot (2x)^2 \cdot 9 + 4 \cdot (2x) \cdot (-27) + 1 \cdot 81

= 16x^4 - 96x^3 + 216x^2 - 216x + 81

Finding Specific Terms

To find the term containing $x^r$ in the expansion of $(a + bx)^n$ :

The general term is $\dbinom{n}{k}a^{n-k}b^k x^k$ .

Set $k = r$ to find the coefficient of $x^r$ .

Find the expansion of $\dfrac{1}{\sqrt{1+x}}$ up to the term in $x^2$ .

(1+x)^\{-1/2\} = 1 + \left(-\frac\{1\}\{2\}\right)x + \frac\{\left(-\frac\{1\}\{2\}\right)\left(-\frac\{3\}\{2\}\right)\}\{2\}x^2

= 1 - \frac\{x\}\{2\} + \frac\{3x^2\}\{8\}

Validity of Expansion

The expansion $(1+x)^n$ for non-integer $n$ converges when $|x| \lt 1$ .

To use this for expressions like $(2 + 3x)^{-1/2}$ Factor out the constant:

(2+3x)^\{-1/2\} = 2^\{-1/2\}\left(1 + \frac\{3x\}\{2\}\right)^\{-1/2\}

This converges when $\left|\dfrac{3x}{2}\right| \lt 1$ I.e., $|x| \lt \dfrac{2}{3}$ .

Mixed Arithmetic-Geometric Sequences

Recurrence Relations

A sequence defined by a recurrence relation gives each term in terms of previous terms.

U_\{n+1\} = au_n + b

Solving First-Order Linear Recurrence Relations

For $u_{n+1} = ru_n + d$ with $u_1$ given:

If $r \neq 1$ The solution is:

U_n = r^\{n-1\}u_1 + d\frac\{r^\{n-1\} - 1\}\{r - 1\}

A sequence is defined by $u_{n+1} = 3u_n + 2$ with $u_1 = 1$ . Find $u_n$ .

$r = 3$ , $d = 2$ .

U_n = 3^\{n-1\} \cdot 1 + 2 \cdot \frac\{3^\{n-1\} - 1\}\{3 - 1\} = 3^\{n-1\} + 3^\{n-1\} - 1 = 2 \cdot 3^\{n-1\} - 1

Verify: $u_1 = 2 \cdot 1 - 1 = 1$ . $u_2 = 2 \cdot 3 - 1 = 5$ . Check: $u_2 = 3(1) + 2 = 5$ .

IB Exam-Style Questions

Question 1 (Paper 1 style)

An arithmetic sequence has first term $5$ and common difference $3$ . A geometric sequence has first Term $3$ and common ratio $2$ . Find the value of $n$ for which the $n$ -th terms are equal.

U_n = 5 + (n-1) \times 3 = 3n + 2

V_n = 3 \times 2^\{n-1\}

3n + 2 = 3 \times 2^\{n-1\}

Testing values:

$n = 1$ : $5 \neq 3$ .

$n = 2$ : $8 \neq 6$ .

$n = 3$ : $11 \neq 12$ .

$n = 4$ : $14 \neq 24$ .

$n = 5$ : $17 \neq 48$ .

The sequences do not have equal terms for small $n$ . The geometric sequence grows much faster. The only possible solution is $n = 3$ approximately (left $= 11$ Right $= 12$ ), so there is no exact Integer solution.

Question 2 (Paper 2 style)

The sum of the first three terms of a geometric sequence is $52$ . The sum of the first six terms is $4732$ . Find the common ratio and the first term.

S_3 = \frac\{u_1(r^3 - 1)\}\{r - 1\} = 52 \quad \mathrm\{(1)\}

S_6 = \frac\{u_1(r^6 - 1)\}\{r - 1\} = 4732 \quad \mathrm\{(2)\}

Dividing (2) by (1):

\frac\{r^6 - 1\}\{r^3 - 1\} = \frac\{4732\}\{52\} = 91

R^3 + 1 = 91 \implies r^3 = 90

Since $90 = 9 \times 10$ is not a perfect cube, this suggests the ratio might not be exact. Let us Reconsider with $r^3 - 1 \neq 0$ :

\frac\{(r^3 - 1)(r^3 + 1)\}\{r^3 - 1\} = r^3 + 1 = 91

R^3 = 90, \quad r = \sqrt[3]\{90\}

From (1): $u_1 = \dfrac{52(r-1)}{r^3 - 1} = \dfrac{52(r-1)}{89}$ .

Question 3 (Paper 1 style)

Expand $(1 - 2x)^5$ in ascending powers of $x$ up to and including the term in $x^3$ .

(1-2x)^5 = 1 + 5(-2x) + 10(-2x)^2 + 10(-2x)^3 + 5(-2x)^4 + (-2x)^5

= 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5

Question 4 (Paper 2 style)

The first three terms of a geometric sequence are $\sin\theta$ , $\cos\theta$ And $\dfrac{1}{\sqrt{3}}$ Where $0 \lt \theta \lt \dfrac{\pi}{2}$ .

Find the value of $\theta$ .

Common ratio: $r = \dfrac{\cos\theta}{\sin\theta} = \cot\theta$ .

Also: $r = \dfrac{1/\sqrt{3}}{\cos\theta}$ .

\cot\theta = \frac\{1\}\{\sqrt\{3\}\cos\theta\}

\frac\{\cos\theta\}\{\sin\theta\} = \frac\{1\}\{\sqrt\{3\}\cos\theta\}

\cos^2\theta = \frac\{\sin\theta\}\{\sqrt\{3\}\}

Since $\cos^2\theta = 1 - \sin^2\theta$ :

1 - \sin^2\theta = \frac\{\sin\theta\}\{\sqrt\{3\}\}

\sqrt\{3\}\sin^2\theta + \sin\theta - \sqrt\{3\} = 0

Let $u = \sin\theta$ :

\sqrt\{3\}u^2 + u - \sqrt\{3\} = 0

U = \frac\{-1 \pm \sqrt\{1 + 12\}\}\{2\sqrt\{3\}\} = \frac\{-1 \pm \sqrt\{13\}\}\{2\sqrt\{3\}\}

Since $0 \lt \theta \lt \dfrac{\pi}{2}$ , $\sin\theta \gt 0$ :

\sin\theta = \frac\{-1 + \sqrt\{13\}\}\{2\sqrt\{3\}\}

Question 5 (Paper 1 style)

Evaluate $\displaystyle\sum_{k=1}^{50} (2k + 1)$ .

\sum_\{k=1\}^\{50\}(2k+1) = 2\sum_\{k=1\}^\{50\}k + \sum_\{k=1\}^\{50\}1 = 2 \cdot \frac\{50 \times 51\}\{2\} + 50 = 2550 + 50 = 2600

Question 6 (Paper 2 style)

The series $1 + 2x + 3x^2 + 4x^3 + \cdots$ can be written as $\displaystyle\sum_{n=1}^{\infty} nx^{n-1}$ .

(a) Find the sum to infinity in terms of $x$ for $|x| \lt 1$ .

We know $\displaystyle\sum_{n=0}^{\infty}x^n = \frac{1}{1-x}$ for $|x| \lt 1$ .

Differentiating both sides with respect to $x$ :

\sum_\{n=1\}^\{\infty\}nx^\{n-1\} = \frac\{1\}\{(1-x)^2\}

(b) Hence find the sum to infinity of $\displaystyle\sum_{n=1}^{\infty}\frac{n}{3^n}$ .

Setting $x = \dfrac{1}{3}$ :

\sum_\{n=1\}^\{\infty\}n\left(\frac\{1\}\{3\}\right)^\{n-1\} = \frac\{1\}\{(1 - 1/3)^2\} = \frac\{1\}\{(2/3)^2\} = \frac\{9\}\{4\}

Therefore:

\sum_\{n=1\}^\{\infty\}\frac\{n\}\{3^n\} = \frac\{1\}\{3\}\sum_\{n=1\}^\{\infty\}n\left(\frac\{1\}\{3\}\right)^\{n-1\} = \frac\{1\}\{3\} \cdot \frac\{9\}\{4\} = \frac\{3\}\{4\}

Summary

Topic	Key Formula
Arithmetic $n$ -th term	$u_n = u_1 + (n-1)d$
Arithmetic sum	$S_n = \dfrac{n}{2}[2u_1 + (n-1)d]$
Geometric $n$ -th term	$u_n = u_1 r^{n-1}$
Geometric sum	$S_n = \dfrac{u_1(1-r^n)}{1-r}$
Sum to infinity	$S_{\infty} = \dfrac{u_1}{1-r}$ for $\\| r \\| \lt 1$
Binomial theorem	$(a+b)^n = \displaystyle\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$
Sigma of $i$	$\displaystyle\sum_{i=1}^{n}i = \dfrac{n(n+1)}{2}$
Sigma of $i^2$	$\displaystyle\sum_{i=1}^{n}i^2 = \dfrac{n(n+1)(2n+1)}{6}$

Prove that $\displaystyle\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$ for all $n \ge 1$ .

Base case ( $n = 1$ ): $\displaystyle\sum_{i=1}^{1} i^2 = 1$ and $\dfrac{1 \times 2 \times 3}{6} = 1$ . True.

Inductive hypothesis: Assume $\displaystyle\sum_{i=1}^{k} i^2 = \frac{k(k+1)(2k+1)}{6}$ .

Inductive step: Show for $n = k + 1$ :

\sum_\{i=1\}^\{k+1\} i^2 = \sum_\{i=1\}^\{k\} i^2 + (k+1)^2 = \frac\{k(k+1)(2k+1)\}\{6\} + (k+1)^2

= \frac\{k(k+1)(2k+1) + 6(k+1)^2\}\{6\} = \frac\{(k+1)[k(2k+1) + 6(k+1)]\}\{6\}

= \frac\{(k+1)(2k^2 + k + 6k + 6)\}\{6\} = \frac\{(k+1)(2k^2 + 7k + 6)\}\{6\}

= \frac\{(k+1)(k+2)(2k+3)\}\{6\} = \frac\{(k+1)((k+1)+1)(2(k+1)+1)\}\{6\}

This is the required formula with $n = k + 1$ . The statement is true by induction.

Additional Sigma Notation Problems

Changing the Index

Sometimes it is useful to change the index of summation:

\sum_\{i=1\}^\{n\} a_i = \sum_\{j=0\}^\{n-1\} a_\{j+1\} = \sum_\{k=2\}^\{n+1\} a_\{k-1\}

Telescoping Series

A telescoping series is one where many terms cancel:

\sum_\{k=1\}^\{n\} \frac\{1\}\{k(k+1)\} = \sum_\{k=1\}^\{n\}\left(\frac\{1\}\{k\} - \frac\{1\}\{k+1\}\right) = 1 - \frac\{1\}\{n+1\} = \frac\{n\}\{n+1\}

Arithmetic-Geometric Mean Inequality

Statement

For positive real numbers $a$ and $b$ :

\frac\{a + b\}\{2\} \ge \sqrt`\{ab}`

Equality holds if and only if $a = b$ .

Generalisation

For positive real numbers $a_1, a_2, \ldots, a_n$ :

\frac\{a_1 + a_2 + \cdots + a_n\}\{n\} \ge \sqrt[n]\{a_1 a_2 \cdots a_n\}

Additional Exam-Style Questions

Question 7 (Paper 2 style)

Use the binomial theorem to expand $(1 + x)^4$ and hence evaluate $1.01^4$ to 5 decimal places.

(1+x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4

Set $x = 0.01$ :

1.01^4 = 1 + 4(0.01) + 6(0.0001) + 4(0.000001) + 0.00000001

= 1 + 0.04 + 0.0006 + 0.000004 + 0.00000001 = 1.04060401

To 5 decimal places: $1.04060$ .

Question 8 (Paper 2 style)

The sum of the first $n$ terms of a sequence is $S_n = n^2 + 2n$ . Find an expression for the $n$ -th Term $u_n$ .

U_n = S_n - S_\{n-1\} = (n^2 + 2n) - ((n-1)^2 + 2(n-1))

= n^2 + 2n - (n^2 - 2n + 1 + 2n - 2) = n^2 + 2n - n^2 + 1 = 2n + 1

Check: $u_1 = S_1 = 3 = 2(1) + 1$ . Correct.

This is an arithmetic sequence with first term 3 and common difference 2.

Question 9 (Paper 2 style)

Find the coefficient of $x^4$ in the expansion of $(2 - 3x)^7$ .

The general term is $\dbinom{7}{k}2^{7-k}(-3x)^k$ .

For $x^4$ : $k = 4$ .

Coefficient $= \dbinom{7}{4} \cdot 2^3 \cdot (-3)^4 = 35 \times 8 \times 81 = 22680$ .

Question 10 (Paper 1 style)

A geometric series has first term $a$ and common ratio $r$ . The sum of the first three terms is $28$ and the sum to infinity is $32$ . Find $a$ and $r$ .

A + ar + ar^2 = 28 \quad \mathrm\{(1)\}

\frac\{a\}\{1-r\} = 32 \implies a = 32(1-r) \quad \mathrm\{(2)\}

Substituting (2) into (1):

32(1-r)(1 + r + r^2) = 28

(1-r)(1+r+r^2) = 1 - r^3 = \frac\{28\}\{32\} = \frac\{7\}\{8\}

R^3 = 1 - \frac\{7\}\{8\} = \frac\{1\}\{8\} \implies r = \frac\{1\}\{2\}

A = 32\left(1 - \frac\{1\}\{2\}\right) = 16

Verify: $16 + 8 + 4 = 28$ . And $S_\infty = 16/(1 - 1/2) = 32$ . Correct.

For the A-Level treatment of this topic, see Sequences and Series.

Tip Ready to test your understanding of Sequences and Series? The contains the hardest questions

within the IB specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Sequences and Series with other IB mathematics topics to test synthesis under exam conditions.

See for instructions on self-marking and building a personal test matrix.

Intuition

A sequence is a list of numbers following a rule, like steps on a staircase. Arithmetic sequences add a constant each step — climbing stairs with uniform height. Geometric sequences multiply by a constant each step — like compound interest where your money grows by a percentage. The sum of a geometric series with ratio less than one converges because each new term is a smaller fraction of the previous, like pouring water into a cup that overflows less and less. The binomial theorem is a shortcut for expanding powers of sums, distributing the work across Pascal’s triangle like a well-organised filing system.

Common Pitfalls

Confusing the domain and range of functions, or not considering restrictions (e.g., denominator cannot be zero).
Forgetting to check that solutions satisfy the original equation (especially with squaring both sides or dividing by variables).

Cross-References

Topic	Site	Link
[Sequences and Series]	A-Level	View
[Sequences and Series]	IB	View
[Sequences and Series]	DSE	View

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Sequences and Series