Number & Algebra — IB Math AA

1. Arithmetic & Geometric Sequences

Arithmetic Sequences

An arithmetic sequence has a constant difference $d$ between consecutive terms. If the first term is $u_1$, the $n$-th term and sum of $n$ terms are:

Arithmetic Sequence Formulas

$$u_n = u_1 + (n-1)d$$

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

Geometric Sequences

A geometric sequence has a constant ratio $r$ between consecutive terms. The $n$-th term, sum of $n$ terms, and infinite sum (when $|r| < 1$) are:

Geometric Sequence Formulas

$$u_n = u_1 \cdot r^{n-1}$$

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

$$S_\infty = \frac{u_1}{1 - r}, \quad |r| < 1$$

Key Insight: The infinite sum $S_\infty$ only exists when $|r| < 1$. If $|r| \geq 1$, the series diverges (no finite sum).

2. Exponents & Logarithms

Logarithms are the inverse of exponential functions. The key identity is: $\log_a b = c \iff a^c = b$.

Laws of Logarithms

$\log_a(xy) = \log_a x + \log_a y$
$\log_a\!\left(\dfrac{x}{y}\right) = \log_a x - \log_a y$
$\log_a(x^n) = n\log_a x$
Change of base: $\log_a x = \dfrac{\ln x}{\ln a} = \dfrac{\log x}{\log a}$

To solve exponential equations, take logarithms of both sides:

$$2^x = 5 \implies x\ln 2 = \ln 5 \implies x = \frac{\ln 5}{\ln 2} \approx 2.322$$

3. Binomial Theorem

The binomial theorem expands $(a + b)^n$ for positive integer $n$:

Binomial Theorem

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

where $\displaystyle\binom{n}{r} = \frac{n!}{r!(n-r)!}$

The general term (the $(r+1)$-th term) is: $T_{r+1} = \binom{n}{r} a^{n-r} b^r$

Pascal's Triangle gives the coefficients: row $n$ gives the coefficients for $(a+b)^n$.

4. Proof by Mathematical Induction (HL)

HL Only: Proof by induction is assessed at Higher Level. SL students should understand sequences and series but do not need to write formal proofs.

Proof by induction has two steps:

Base Case: Verify the statement is true for $n = 1$ (or the starting value).
Inductive Step: Assume the statement is true for $n = k$ (inductive hypothesis), then prove it is true for $n = k+1$.

Classic Example: Sum of First n Integers

Prove: $\displaystyle\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$

Base case ($n=1$): LHS $= 1$. RHS $= \frac{1 \cdot 2}{2} = 1$. ✓

Inductive step: Assume true for $n = k$: $\sum_{i=1}^{k} i = \frac{k(k+1)}{2}$.

Then $\sum_{i=1}^{k+1} i = \frac{k(k+1)}{2} + (k+1) = \frac{k(k+1) + 2(k+1)}{2} = \frac{(k+1)(k+2)}{2}$. ✓

This is exactly the formula with $n = k+1$, so by induction the result holds for all $n \geq 1$.

5. Partial Fractions (HL)

HL Only: Partial fraction decomposition is required for HL, particularly for integration.

To decompose a rational function, write it as a sum of simpler fractions:

$$\frac{f(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$$

Multiply both sides by $(x-a)(x-b)$ and substitute $x = a$ and $x = b$ to find $A$ and $B$.

Repeated factor: $\dfrac{f(x)}{(x-a)^2} = \dfrac{A}{x-a} + \dfrac{B}{(x-a)^2}$

Worked Examples

Worked Example 1

Geometric Series — Finding the Sum to Infinity

A geometric sequence has first term $u_1 = 12$ and common ratio $r = \frac{1}{3}$. Find (a) the 5th term, (b) the sum of the first 6 terms, and (c) the sum to infinity.

Part (a): 5th term
$u_5 = 12 \cdot \left(\frac{1}{3}\right)^4 = 12 \cdot \frac{1}{81} = \frac{12}{81} = \frac{4}{27}$

Part (b): Sum of 6 terms
$S_6 = \frac{12(1 - (1/3)^6)}{1 - 1/3} = \frac{12(1 - 1/729)}{2/3} = 18\left(\frac{728}{729}\right) = \frac{13104}{729} = \frac{4368}{243} \approx 17.98$

Part (c): Sum to infinity
Since $|r| = \frac{1}{3} < 1$: $S_\infty = \frac{12}{1 - 1/3} = \frac{12}{2/3} = 18$

Worked Example 2

Binomial Theorem — Finding a Specific Term

Find the term independent of $x$ in the expansion of $\left(2x - \dfrac{1}{x^2}\right)^6$.

Write the general term: $T_{r+1} = \binom{6}{r}(2x)^{6-r}\left(-\frac{1}{x^2}\right)^r = \binom{6}{r} \cdot 2^{6-r} \cdot (-1)^r \cdot x^{6-r} \cdot x^{-2r}$

Collect powers of $x$: $x^{6-r-2r} = x^{6-3r}$. For the term independent of $x$, set $6 - 3r = 0 \implies r = 2$.

Substitute $r = 2$: $T_3 = \binom{6}{2} \cdot 2^4 \cdot (-1)^2 = 15 \cdot 16 \cdot 1 = \boxed{240}$

Worked Example 3

Exponential Equation — Using Logarithms

Solve $3^{2x+1} = 7^{x-1}$. Give your answer to 3 significant figures.

Take $\ln$ of both sides: $(2x+1)\ln 3 = (x-1)\ln 7$

Expand: $2x\ln 3 + \ln 3 = x\ln 7 - \ln 7$

Collect $x$ terms: $x(2\ln 3 - \ln 7) = -\ln 7 - \ln 3$

Solve: $x = \dfrac{-\ln 7 - \ln 3}{2\ln 3 - \ln 7} = \dfrac{-\ln 21}{2\ln 3 - \ln 7} \approx \dfrac{-3.045}{-0.249} \approx \boxed{12.2}$

Practice Problems

Q1. An arithmetic sequence has $u_3 = 11$ and $u_8 = 26$. Find the first term and common difference. Hence find $S_{20}$.

Show Solution

From the two conditions: $u_1 + 2d = 11$ and $u_1 + 7d = 26$.

Subtracting: $5d = 15 \implies d = 3$. Then $u_1 = 11 - 6 = 5$.

$S_{20} = \frac{20}{2}(2(5) + 19(3)) = 10(10 + 57) = 10 \times 67 = \mathbf{670}$

Q2. The 2nd term of a geometric sequence is 6 and the 5th term is $\frac{6}{125}$. Find the common ratio and the sum to infinity.

Show Solution

$u_2 = u_1 r = 6$ and $u_5 = u_1 r^4 = \frac{6}{125}$.

Dividing: $r^3 = \frac{6/125}{6} = \frac{1}{125} \implies r = \frac{1}{5}$.

Then $u_1 = 6/r = 30$. Since $|r| = 1/5 < 1$: $S_\infty = \frac{30}{1 - 1/5} = \frac{30}{4/5} = \mathbf{37.5}$

Q3. Expand $(1 + 3x)^5$ up to and including the term in $x^3$.

Show Solution

Using the binomial theorem with $a=1$, $b=3x$, $n=5$:

$T_1 = \binom{5}{0}(3x)^0 = 1$

$T_2 = \binom{5}{1}(3x)^1 = 15x$

$T_3 = \binom{5}{2}(3x)^2 = 10 \cdot 9x^2 = 90x^2$

$T_4 = \binom{5}{3}(3x)^3 = 10 \cdot 27x^3 = 270x^3$

$(1+3x)^5 \approx 1 + 15x + 90x^2 + 270x^3 + \cdots$

Q4 (HL). Decompose $\dfrac{5x+1}{(x+1)(x-2)}$ into partial fractions.

Show Solution

Write $\frac{5x+1}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2}$.

Multiply through: $5x+1 = A(x-2) + B(x+1)$.

Let $x = 2$: $11 = 3B \implies B = \frac{11}{3}$.

Let $x = -1$: $-4 = -3A \implies A = \frac{4}{3}$.

$$\frac{5x+1}{(x+1)(x-2)} = \frac{4/3}{x+1} + \frac{11/3}{x-2}$$