Number & Algebra

An ordered list of numbers $u_1, u_2, u_3, \dots$ following a rule. Can be finite or infinite.

The sum of a sequence: $u_1 + u_2 + \dots + u_n$. Finite (sum to $u_n$) or infinite (sum forever).

$u_n$ is the $n$-th term, expressed as a formula in $n$. Two sequences differ iff their general terms do.

Common difference $d$ is constant across consecutive terms.

$u_p - u_q = (p - q)d$ — the gap is exactly $(p-q)$ common differences.

Common ratio $r$ is constant. $u_n \neq 0$ required.

Convergent if $|r|<1$, divergent otherwise. Always check before applying.

Use when given $S_n$ as a formula in $n$.

$r\%$ per year, $n$ years; interest stays linear.

$k$ = compounding periods per year. Per-period rate = $r/(100k)$; periods = $kn$.

Same shape; $r < 0$ for depreciation.

Pair the 1st and last term, 2nd and 2nd-last, … each pair sums to $u_1+u_n$. With $n$ terms there are $n/2$ pairs → derives $S_n = \tfrac{n}{2}(u_1+u_n)$.

Use $u_p - u_q = (p-q)d$ to solve for $d$ in ONE step.

Divide them: $u_p/u_q = r^{p-q}$. Take the appropriate root.

$|r| = 1$ does NOT converge (Grandi's series $1-1+1-1+\dots$ is the classic warning). Always check $|r| < 1$ strictly.

If interest compounds $k$ times per year, the EXPONENT is $kn$ (not $n$); the per-period rate is $r/(100k)$, not $r/100$.

A valid step-by-step argument that establishes the truth of a mathematical statement. Examples DON'T constitute proof — they merely suggest.

Cautionary: $f(n)=n^2+n+41$ gives primes for $n = 0, \dots, 39$ — then fails at $n=40$ where $f(40)=41^2$. Patterns that hold for many cases can still be false.

Start from what's given; use algebra + known facts; reach the conclusion in steps with no leaps. Mark the end with $\blacksquare$.

Sum of two evens is even. Template: 'Let $m, n \in \mathbb{Z}$. Then $2m, 2n$ are even. Their sum is $2(m+n) \in 2\mathbb{Z}$. $\blacksquare$'

Building blocks for nearly every algebraic direct proof.

Used constantly in number-theory proofs.

To disprove a 'for all' statement. ONE failing case is enough; no need to find more.

'Every prime is odd' — counterexample: $n=2$ is prime AND even. Claim disproved.

Counterexamples DISPROVE universal claims only. They can't disprove existence claims like 'there exists $x$ with …'.

(1) Assume the negation of the statement. (2) Derive a logical impossibility or contradiction with known fact. (3) Conclude the original must be true. $\blacksquare$

Assume $\sqrt 2 = p/q$ in lowest terms. Square: $p^2 = 2q^2 \Rightarrow p$ even. Then $4k^2 = 2q^2 \Rightarrow q$ also even. But $p, q$ share factor 2 — contradicts 'lowest terms'.

Assume primes finite: $p_1, \dots, p_n$. Let $N = p_1 p_2 \cdots p_n + 1$. Then $N$ has a prime divisor distinct from all $p_i$ — contradiction.

(1) Base case: show $P(1)$ (or $P(n_0)$ where the claim starts). (2) Inductive step: assume $P(k)$, prove $P(k+1)$. Conclude $P(n)$ for all $n \ge n_0$. $\blacksquare$

Base case knocks the first domino. Inductive step says each falling domino knocks the next. Conclude: all fall. Both parts essential — omit either and the chain breaks.

The assumed statement $P(k)$. To bridge to $P(k+1)$: add the $(k+1)$-th term to both sides (for sums), or multiply both sides by the relevant factor.

Base: $1 = 1\cdot2/2$ ✓. Step: add $(k+1)$ to IH ⇒ $\tfrac{k(k+1)}{2} + (k+1) = \tfrac{(k+1)(k+2)}{2}$ ✓.

Prove $2^n > n^2$ for $n \ge 5$. Base $n=5$: $32 > 25$ ✓. Step: $2^{k+1} = 2\cdot 2^k > 2k^2 \ge (k+1)^2$ since $2k^2-(k+1)^2 = k^2-2k-1 > 0$ for $k \ge 5$.

'For all $n \in \mathbb{Z}^+$' → induction. 'For all $x \in \mathbb{R}$, $P$' → direct proof. 'For all …' suspected false → counterexample. 'No such $x$' or '$x$ is irrational' → contradiction.

You MUST use the inductive hypothesis $P(k)$ inside the proof of $P(k+1)$. If the hypothesis is never invoked, your 'proof' isn't induction — it's just direct.

Make sure the negation is fully derived to a CONTRADICTION (e.g. '$a$ both even and odd', '$0 = 1$'). Just 'this doesn't look right' is not enough.

A single example does NOT prove a 'for all' claim — it only suggests. Don't try to 'prove' by finding one positive case.

If task A has $a$ outcomes and task B has $b$ outcomes, doing both in sequence gives $a \times b$ outcomes. Generalises to any number of independent stages.

If task A has $a$ outcomes and task B has $b$ outcomes and they are mutually exclusive alternatives, choosing one or the other gives $a + b$ outcomes.

$1!=1,\;2!=2,\;3!=6,\;4!=24,\;5!=120,\;6!=720,\;7!=5040$. Grows extremely fast — by $10!$ you are past 3.6 million.

Number of orderings of $n$ different items along a line.

Pick $r$ items from $n$ AND arrange them. Example: 3-letter codes from 5 letters with no repeats $= {}^5P_3 = 60$.

Arrange $n$ items where $n_1, n_2, \dots$ are repeats of each kind. Example: arrangements of MISSISSIPPI $= \tfrac{11!}{1!\,4!\,4!\,2!} = 34{,}650$.

Round table: fix one person to remove rotational symmetry, then arrange the remaining $n-1$ linearly.

Pick $r$ items from $n$ when order is irrelevant. Equal to $\tfrac{{}^nP_r}{r!}$ — divide out the $r!$ orderings of each chosen group.

Choosing $r$ to include = choosing $n-r$ to leave out.

Each entry of Pascal's triangle is the sum of the two above it. Useful for shortcuts and proofs.

Expands $(a+b)^n$ into $n+1$ terms: powers of $a$ fall from $n$ to $0$, powers of $b$ rise from $0$ to $n$, coefficients come from row $n$ of Pascal's triangle.

The $(r+1)$-th term, indexed from $r=0$. Use this to extract a specific term (e.g. the $x^5$ term) without expanding the whole thing.

To find the coefficient of $x^k$ in $(\alpha + \beta x)^n$: write $T_{r+1} = \binom{n}{r}\alpha^{n-r}(\beta x)^{r}$, set the power of $x$ to $k$, solve for $r$, evaluate.

Row 0: 1 — Row 1: 1 1 — Row 2: 1 2 1 — Row 3: 1 3 3 1 — Row 4: 1 4 6 4 1 — Row 5: 1 5 10 10 5 1 — Row 6: 1 6 15 20 15 6 1.

Ask: does the order of the chosen items change the outcome? YES ⇒ permutation $({}^nP_r)$. NO ⇒ combination $({}^nC_r)$. Seating, codes, ordered podiums ⇒ P. Committees, hands of cards, picking a team ⇒ C.

AND between independent stages ⇒ multiply. OR between mutually exclusive cases ⇒ add. Split a mixed problem into disjoint cases first (add), then count each case (multiply).

If swapping two identical items gives the same arrangement, you have over-counted. Divide by the symmetries (use the identical-objects formula or factor out the equivalent orderings).

In $(a + bx)^n$, every term carries an extra power of $b$ as well as $x$. Do not drop the $b^r$. In $(a - bx)^n$, term $r$ carries $(-1)^r$.

${}^nP_r$ and ${}^nC_r$ have dedicated GDC buttons (often under MATH → PRB). Do not compute $n!/(n-r)!$ for large $n$ — overflow risk. Use the built-in function.

Mixed practice over sequences/series, proof, counting & binomial.

18 questions across 4 sections (A: 6, B: 3, C: 5, D: 4 multi-part). Designed for ~60 min in one sitting; pause and resume any time.

Use this to recover $d$ when two non-consecutive terms are given.

$i$ = period rate (annual rate $\div$ compounds per year). $n$ = total periods. Depreciation: same formula with $1 - i$.

Express assumed even/odd integers this way, then manipulate.

Template: write each as $2k$, factor 2 out, conclude.

ONE concrete instance is enough to disprove a 'for all' claim. Example: $n^{2}+n+41$ is prime for $n=0,1,\dots,39$ but $n=40$ gives $41^{2}$ — not prime.

Assume the OPPOSITE of what you want, derive a logical impossibility (e.g. an integer that is both even and odd), conclude the original must hold.

Assume $\sqrt{2}=p/q$ in lowest terms ⇒ $p^{2}=2q^{2}$ ⇒ $p$ even ⇒ $q$ even ⇒ contradicts 'lowest terms'.

Base case, inductive hypothesis (IH), inductive step, conclusion. The IH must appear in the step.

Step: add $(k+1)$ to IH ⇒ $\tfrac{k(k+1)}{2}+(k+1) = \tfrac{(k+1)(k+2)}{2}$.

'For all $n\in\mathbb{Z}^{+}$' → induction. 'For all $x\in\mathbb{R}$, $P$' → direct. suspected false → counter-example. '$x$ irrational' or 'no such $x$' → contradiction.

Independent stages with $a$, $b$ outcomes give $a \times b$ total.

Mutually exclusive alternatives with $a$, $b$ outcomes give $a + b$ total.

Fix one seat to kill rotational symmetry.

Indexed from $r=0$. Use to pick out a specific term ($x^{k}$, constant term, …).

For $(\alpha + \beta x)^{n}$: write $T_{r+1}=\binom{n}{r}\alpha^{n-r}(\beta x)^{r}$, set the power of $x$ equal to the target, solve for $r$, evaluate.

Pick up $(-1)^{r}$ in each term.

$S_{\infty}$ exists ONLY for $|r|<1$ strict. $r=\pm 1$ does NOT converge (Grandi's series).

If interest is compounded $m$ times/year for $t$ years, $n = m\,t$ and the periodic rate is $\tfrac{\text{annual}}{m}$.

In $(a+bx)^{n}$, each term carries an extra $b^{r}$ as well as $x^{r}$ — don't drop the $b$.

You MUST use the IH inside the step. If the hypothesis $P(k)$ is never invoked, it's not induction — it's just a direct proof for $n=k+1$.

HL only. 2 investigations, ~55 marks total, 1 hour. Each problem builds 8-12 parts from compute → conjecture → prove. Calculator allowed.

~30 min per investigation. Don't get stuck on one part — move to the next; later parts may give you clues for earlier ones.

Almost every part: M1 (method shown) + A1 (correct value/expression) + R1 (justification or conclusion). The R1 marks compound — losing them kills your top band.

$z_{n+1} = f(z_n)$ — fixed points satisfy $z = f(z)$; long-term behaviour ⇒ attracting (sequence converges to it) or repelling.