Sequences and Series

Geometric Series Convergence

Example

Find the 20th term and the sum of the first 20 terms of the sequence $3, 7, 11, 15, \ldots$.

$u_1 = 3$, $d = 4$.

Properties of Arithmetic Sequences

The arithmetic mean of $a$ and $b$ is $\dfrac{a+b}{2}$.

If $a, b, c$ are consecutive terms of an arithmetic sequence, then $2b = a + c$.

Example

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

(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:

General Term

Sum of the First $n$ Terms

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

Example

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

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

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:

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

Example

Find the sum to infinity of $1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots$.

$u_1 = 1$, $r = \dfrac{1}{2}$. Since $|r| \lt 1$:

Example

Express $0.\dot{7}$ (recurring decimal) as a fraction.

This is a geometric series with $u_1 = \dfrac{7}{10}$ and $r = \dfrac{1}{10}$.

Example

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

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

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

Properties

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

Useful Sigma Summations

Example

Evaluate $\displaystyle\sum_{k=1}^{10} (3k - 1)$.

Example

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

Applications

Compound Interest

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

Population Growth

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

Depreciation

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

Example

USD 5000 is invested at 6% per year, compounded annually. Find the value after 10 years.

The investment is worth approximately USD 8954.24.

Example

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

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:

The Binomial Theorem

Expansion of $(a + b)^n$

For positive integer $n$:

Where the binomial coefficient is:

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:

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

Example

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

Using the binomial theorem:

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 $x^r$.

Example

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

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

For $x^3$: $k = 3$.

Coefficient $= \dbinom{7}{3} \cdot 2^4 \cdot 3^3 = 35 \times 16 \times 27 = 15120$.

Binomial Expansion for Negative or Fractional Powers

For $|x| \lt 1$ and $n \in \mathbb{Q}$:

This is an infinite series that converges for $|x| \lt 1$.

Example

Find the expansion of $(1 + x)^{-2}$ up to the term in $x^3$.

Example

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

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:

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.

Solving First-Order Linear Recurrence Relations

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

If $r \neq 1$The solution is:

Example

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

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

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.

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.

Dividing (2) by (1):

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

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 $x^3$.

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

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

Let $u = \sin\theta$:

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

Question 5 (Paper 1 style)

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

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 $x$:

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

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

Therefore:

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 $i^2$ | $\displaystyle\sum_{i=1}^{n}i^2 = \dfrac{n(n+1)(2n+1)}{6}$ | | |

Example

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 $n = k + 1$:

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:

Telescoping Series

A telescoping series is one where many terms cancel:

Example

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

Using partial fractions:

Arithmetic-Geometric Mean Inequality

Statement

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

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

Generalisation

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

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.

Set $x = 0.01$:

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 $u_n$.

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 $r$.

Substituting (2) into (1):

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 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.

Common Pitfalls

1. Confusing the domain and range of functions, or not considering restrictions (e.g., denominator cannot be zero).

2. Forgetting to check that solutions satisfy the original equation (especially with squaring both sides or dividing by variables).

Cross-References

Worked Examples

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