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SLIBMathematics

Mathematics: Analysis and Approaches

The Standard Level pathway develops mathematical reasoning through functions, calculus, statistics, and proof.

Course units

5

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Course Overview

IB Mathematics: Analysis and Approaches SL is organized into 5 units and 51 focused topics with 737 Key Concepts sections, 792 questions, and 1020 flashcards.

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Begin with Topic 1.1: Scientific notation and operations with numbers.

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Course units

Each unit groups focused topics. Every topic follows the same clear rhythm: review Key Concepts, use the Question Bank, then strengthen recall.

  1. Unit 1
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    Number and Algebra

    This topic builds the algebraic language used throughout AA SL. Sequences, exponents, logarithms, proof and the binomial theorem later reappear inside functions, probability and calculus.

    9 topics

    Unit 1 progress0%

    Topics in this unit

    Topic 1.1Not started

    Scientific notation and operations with numbers

    Scientific notation is a compact, reliable way to represent, compare and calculate with extremely large or small numbers. This guide builds from place value and conversions to arithmetic, significant figures, calculator use and IB-style multi-step problems.

    105 min estimated

    Topic 1.1 progress0%
    Learning objectives
    • Write every non-zero number in normalized scientific notation a × 10k10^{k}, where 1 ≤ |a| < 10 and k is an integer.
    • Convert accurately between ordinary decimal notation, scientific notation and calculator E-notation.
    • Multiply, divide and raise scientific-notation values to integer powers using exponent laws.
    • Add and subtract scientific-notation values by first expressing them with a common power of ten.
    • Round values to a stated number of significant figures and communicate trailing significant zeros clearly.
    • Use a scientific calculator efficiently while retaining guard digits until the final answer.
    • Estimate orders of magnitude and use them to check whether a calculated result is reasonable.
    • Present IB exam solutions with correct working, notation, accuracy and units.
    Key formulas

    Normalized scientific notation

    N=a×10k,1≤∣a∣<10,k∈ZN=a\times10^k,\quad 1\le |a|<10,\quad k\in\mathbb Z

    Multiplication rule

    (a×10m)(b×10n)=(ab)×10m+n(a\times10^m)(b\times10^n)=(ab)\times10^{m+n}

    Division rule

    a×10mb×10n=ab×10m−n,b≠0\frac{a\times10^m}{b\times10^n}=\frac{a}{b}\times10^{m-n},\quad b\ne0

    Power rule

    (a×10m)p=ap×10mp,p∈Z,a≠0(a\times10^m)^p=a^p\times10^{mp},\quad p\in\mathbb Z,\quad a\ne0

    Addition with a common exponent

    a×10k+b×10k=(a+b)×10ka\times10^k+b\times10^k=(a+b)\times10^k

    Subtraction with a common exponent

    a×10k−b×10k=(a−b)×10ka\times10^k-b\times10^k=(a-b)\times10^k

    Calculator E-notation

    aEk=a×10ka\text{E}k=a\times10^k
    Topic 1.2Not started

    Arithmetic sequences and series

    Arithmetic sequences model quantities that change by the same additive amount each step. This guide develops common differences, nth terms, finite sums, sigma notation and IB-style applications from first principles.

    110 min estimated

    Topic 1.2 progress0%
    Learning objectives
    • Recognize an arithmetic sequence and calculate its common difference from consecutive or non-consecutive terms.
    • Use recurrence notation and the nth-term formula to generate and find terms.
    • Determine an unknown first term, common difference, term number or number of terms from given information.
    • Distinguish an arithmetic sequence from an arithmetic series and calculate the sum of a finite number of terms.
    • Represent and evaluate finite arithmetic sums using sigma notation.
    • Build and interpret arithmetic models for real-world situations with constant additive change.
    • Use calculator tables, summation tools and equation solvers efficiently while checking integer constraints.
    • Present IB exam solutions with clear formulas, substitutions, units and reasonableness checks.
    Key formulas

    Common difference

    d=un+1−und=u_{n+1}-u_n

    Recursive definition

    u1=a,un+1=un+du_1=a,\qquad u_{n+1}=u_n+d

    Arithmetic nth term

    un=u1+(n−1)du_n=u_1+(n-1)d

    Common difference from two terms

    d=uq−upq−p,p≠qd=\frac{u_q-u_p}{q-p},\qquad p\ne q

    Term number

    n=1+un−u1d,d≠0n=1+\frac{u_n-u_1}{d},\qquad d\ne0

    Arithmetic series using first and last terms

    Sn=n2(u1+un)S_n=\frac n2(u_1+u_n)

    Arithmetic series using the common difference

    Sn=n2[2u1+(n−1)d]S_n=\frac n2\left[2u_1+(n-1)d\right]

    Arithmetic series in sigma notation

    Sn=∑r=1n[u1+(r−1)d]S_n=\sum\limits_{r=1}^{n}\left[u_1+(r-1)d\right]

    Equivalent zero-based indexing

    Sn=∑r=0n−1(u1+rd)S_n=\sum\limits_{r=0}^{n-1}(u_1+rd)
    Topic 1.3Not started

    Geometric sequences and series

    Geometric sequences model repeated multiplication. This guide develops common ratios, nth terms, finite and infinite sums, convergence, sigma notation and IB-style growth and decay applications.

    120 min estimated

    Topic 1.3 progress0%
    Learning objectives
    • Recognize a geometric sequence and calculate its common ratio, including fractional and negative ratios.
    • Use recurrence notation and the geometric nth-term formula to generate and find terms.
    • Determine an unknown first term, common ratio, term number or number of terms from given information.
    • Distinguish a geometric sequence from a geometric series and calculate finite sums.
    • Decide whether an infinite geometric series converges and calculate its sum when it exists.
    • Represent and evaluate finite and infinite geometric sums using sigma notation.
    • Build and interpret repeated-percentage growth and decay models in context.
    • Use calculator tables, logarithms and numerical solvers while respecting domain and integer constraints.
    • Present IB solutions with clear formulas, substitutions, units, accuracy and reasonableness checks.
    Key formulas

    Common ratio

    r=un+1un,un≠0r=\frac{u_{n+1}}{u_n},\qquad u_n\ne0

    Recursive definition

    u1=a,un+1=runu_1=a,\qquad u_{n+1}=ru_n

    Geometric nth term

    un=u1rn−1u_n=u_1r^{n-1}

    Ratio from two terms

    rq−p=uqupr^{q-p}=\frac{u_q}{u_p}

    Term number for a positive ratio

    n=1+ln⁡(unu1)ln⁡r,r>0, r≠1,u1≠0,unu1>0n=1+\frac{\ln(\frac{u_n}{u_1})}{\ln r},\qquad r>0,\ r\ne1,\quad u_1\ne0,\quad \frac{u_n}{u_1}>0

    Finite geometric sum

    Sn=u1(1−rn)1−r,r≠1S_n=\frac{u_1(1-r^n)}{1-r},\qquad r\ne1

    Equivalent finite-sum form

    Sn=u1(rn−1)r−1,r≠1S_n=\frac{u_1(r^n-1)}{r-1},\qquad r\ne1

    Special case \(r=1\)

    Sn=nu1S_n=nu_1

    Infinite geometric sum

    S∞=u11−r,∣r∣<1S_\infty=\frac{u_1}{1-r},\qquad |r|<1

    Remainder after n terms

    Rn=S∞−Sn=u1rn1−r,∣r∣<1R_n=S_\infty-S_n=\frac{u_1r^n}{1-r},\qquad |r|<1

    Finite geometric series in sigma notation

    Sn=∑k=0n−1u1rkS_n=\sum\limits_{k=0}^{n-1}u_1r^k

    Infinite geometric series in sigma notation

    ∑k=0∞u1rk=u11−r,∣r∣<1\sum\limits_{k=0}^{\infty}u_1r^k=\frac{u_1}{1-r},\qquad |r|<1

    Repeated growth or decay

    Qt=Q0rtQ_t=Q_0r^t
    Topic 1.4Not started

    Financial applications

    Financial mathematics models how money changes through simple interest, compound growth, depreciation, regular savings and loan repayments. Accurate solutions depend on matching the interest rate to the compounding or payment period.

    140 min estimated

    Topic 1.4 progress0%
    Learning objectives
    • Distinguish simple interest from compound interest and select the appropriate model.
    • Convert percentage rates to decimal multipliers for growth, appreciation, decay and depreciation.
    • Calculate future value, present value, interest rate or time using compound-interest models.
    • Match nominal annual rates, periodic rates, compounding frequency and elapsed time consistently.
    • Calculate and interpret effective annual interest rates.
    • Model regular savings deposits using finite geometric-series formulas.
    • Model amortizing loans, calculate periodic repayments and interpret outstanding balances.
    • Use GDC financial or TVM functions with a consistent cash-flow sign convention.
    • Solve financial word problems with appropriate units, accuracy, integer constraints and contextual interpretation.
    • Evaluate whether a financial model and its assumptions are reasonable in a real situation.
    Key formulas

    Simple interest earned

    I=PrtI=Prt

    Simple-interest balance

    A=P(1+rt)A=P(1+rt)

    Percentage multiplier

    growth: 1+p100,decay: 1−p100\text{growth: }1+\frac{p}{100},\qquad \text{decay: }1-\frac{p}{100}

    Compound future value

    FV=PV(1+i)nFV=PV(1+i)^n

    Periodic rate from two values

    i=(FVPV)1n−1,n>0, PV≠0, FVPV>0i=\left(\frac{FV}{PV}\right)^{\frac{1}{n}}-1,\quad n>0,\ PV\ne0,\frac{\ FV}{PV}>0

    Number of periods

    n=ln⁡(FV/PV)ln⁡(1+i)PV≠0,FV/PV>0,i>−1, i≠0\begin{gathered}n=\frac{\ln(FV/PV)}{\ln(1+i)}\\ PV\ne0,\quad FV/PV>0,\quad i>-1,\ i\ne0\end{gathered}

    Nominal-rate compound interest

    FV=PV(1+rm)mtFV=PV\left(1+\frac{r}{m}\right)^{mt}

    Periodic rate and period count

    i=rm,n=mti=\frac{r}{m},\qquad n=mt

    Effective annual rate

    EAR=(1+rm)m−1EAR=\left(1+\frac{r}{m}\right)^m-1

    Present value of one future amount

    PV=FV(1+i)n=FV(1+i)−nPV=\frac{FV}{(1+i)^n}=FV(1+i)^{-n}

    Appreciation

    Vn=V0(1+a)nV_n=V_0(1+a)^n

    Reducing-balance depreciation

    Vn=V0(1−d)nV_n=V_0(1-d)^n

    Future value of an ordinary annuity

    FV={R(1+i)n−1i,i≠0nR,i=0FV=\begin{cases}R\frac{(1+i)^n-1}{i},&i\ne0\\ nR,&i=0\end{cases}

    Future value of an annuity due

    FVdue={R(1+i)n−1i(1+i),i≠0nR,i=0FV_{\mathrm{due}}=\begin{cases}R\frac{(1+i)^n-1}{i}(1+i),&i\ne0\\ nR,&i=0\end{cases}

    Deposit required for a target

    R={FVi(1+i)n−1,i≠0FV/n,i=0R=\begin{cases}FV\frac{i}{(1+i)^n-1},&i\ne0\\ FV/n,&i=0\end{cases}

    Present value of loan repayments

    PV={R1−(1+i)−ni,i≠0nR,i=0PV=\begin{cases}R\frac{1-(1+i)^{-n}}{i},&i\ne0\\ nR,&i=0\end{cases}

    Periodic loan repayment

    R={PVi1−(1+i)−n,i≠0PV/n,i=0R=\begin{cases}PV\frac{i}{1-(1+i)^{-n}},&i\ne0\\ PV/n,&i=0\end{cases}

    Outstanding balance after k payments

    Bk={PV(1+i)k−R(1+i)k−1i,i≠0PV−kR,i=0B_k=\begin{cases}PV(1+i)^k-R\frac{(1+i)^k-1}{i},&i\ne0\\ PV-kR,&i=0\end{cases}

    Total interest on equal repayments

    Itotal=Rn−PVI_{\mathrm{total}}=Rn-PV
    Topic 1.5Not started

    Integer exponents and introduction to logarithms

    Integer exponents describe repeated multiplication and reciprocal powers, while logarithms reverse exponentiation. This guide develops the exponent laws, logarithmic notation, exact and numerical equation solving, calculator methods, and IB exam reasoning.

    125 min estimated

    Topic 1.5 progress0%
    Learning objectives
    • Interpret positive, zero, and negative integer exponents and state the restrictions that make each expression defined.
    • Apply the product, quotient, and power laws accurately when simplifying algebraic expressions.
    • Rewrite answers with positive exponents when required and distinguish powers from ordinary multiplication.
    • Explain a logarithm as an exponent and convert fluently between exponential and logarithmic form.
    • Use the basic product, quotient, and power properties of logarithms with valid positive arguments.
    • Solve simple exponential equations exactly when a common base is available and numerically using logarithms otherwise.
    • Solve simple logarithmic equations while enforcing domain and base restrictions.
    • Use a scientific calculator or GDC efficiently and communicate IB solutions with appropriate working and accuracy.
    Key formulas

    Positive integer exponent

    an=a⋅a⋯a⏟n factors,n∈Na^n=\underbrace{a\cdot a\cdots a}_{n\text{ factors}},\quad n\in\mathbb N

    Zero exponent

    a0=1,a≠0a^0=1,\quad a\ne0

    Negative integer exponent

    a−n=1an,a≠0a^{-n}=\frac{1}{a^n},\quad a\ne0

    Product law

    aman=am+na^m a^n=a^{m+n}

    Quotient law

    aman=am−n,a≠0\frac{a^m}{a^n}=a^{m-n},\quad a\ne0

    Power of a power

    (am)n=amn(a^m)^n=a^{mn}

    Power of a product

    (ab)n=anbn(ab)^n=a^n b^n

    Power of a quotient

    (ab)n=anbn,b≠0\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n},\quad b\ne0

    Moving a negative power across a fraction bar

    a−mb−n=bnam,a,b≠0\frac{a^{-m}}{b^{-n}}=\frac{b^n}{a^m},\quad a,b\ne0

    Exponential-logarithmic equivalence

    ac=b⟺log⁡ab=ca^c=b\quad\Longleftrightarrow\quad\log_a b=c

    Common and natural logarithms

    log⁡x=log⁡10x,ln⁡x=log⁡ex\log x=\log_{10}x,\qquad\ln x=\log_e x

    Inverse identities

    log⁡a(ax)=x,alog⁡ax=x(x>0)\log_a(a^x)=x,\qquad a^{\log_a x}=x\quad(x>0)

    Product property

    log⁡a(xy)=log⁡ax+log⁡ay\log_a(xy)=\log_a x+\log_a y

    Quotient property

    log⁡a(xy)=log⁡ax−log⁡ay\log_a\left(\frac{x}{y}\right)=\log_a x-\log_a y

    Power property

    log⁡a(xk)=klog⁡ax\log_a(x^k)=k\log_a x

    Useful exact values

    log⁡a1=0,log⁡aa=1\log_a1=0,\qquad\log_a a=1

    Common-base principle

    af(x)=ag(x)⟹f(x)=g(x),a>0, a≠1a^{f(x)}=a^{g(x)}\Longrightarrow f(x)=g(x),\quad a>0,\ a\ne1

    Solving a basic exponential equation

    ax=b⟹x=ln⁡bln⁡a,a>0, a≠1, b>0a^x=b\Longrightarrow x=\frac{\ln b}{\ln a},\quad a>0,\ a\ne1,\ b>0

    Single logarithm equation

    log⁡af(x)=c⟺f(x)=ac,f(x)>0\log_a f(x)=c\Longleftrightarrow f(x)=a^c,\quad f(x)>0

    Equality of logarithms

    log⁡af(x)=log⁡ag(x)⟹f(x)=g(x),f(x),g(x)>0\log_a f(x)=\log_a g(x)\Longrightarrow f(x)=g(x),\quad f(x),g(x)>0

    Change-of-base formula

    log⁡ab=log⁡blog⁡a=ln⁡bln⁡a\log_a b=\frac{\log b}{\log a}=\frac{\ln b}{\ln a}
    Topic 1.6Not started

    Simple deductive proof

    Deductive proof turns definitions and known facts into a logically certain conclusion. This guide develops direct proof, algebraic proof, parity, divisibility, consecutive-integer arguments, and the ability to distinguish proof from unsupported pattern spotting.

    115 min estimated

    Topic 1.6 progress0%
    Learning objectives
    • Explain deductive proof as a finite chain of justified statements that establishes a conclusion for every permitted case.
    • Identify hypotheses, conclusions, implications, universal statements, identities, and counterexamples.
    • Use mathematical definitions and previously established results explicitly rather than relying on numerical evidence.
    • Write a clear direct proof with a stated starting assumption and a conclusion that answers the original claim.
    • Prove algebraic identities by transforming one side into the other through valid reversible steps.
    • Represent even, odd, divisible, and consecutive integers algebraically and use those representations in proof.
    • Determine whether an argument is valid, detect hidden assumptions or circular reasoning, and explain why a proposed proof fails.
    • Communicate concise IB-style proofs using correct notation, complete sentences, and appropriate quantifiers.
    Key formulas

    Implication structure

    P⇒QP\Rightarrow Q

    Universal statement

    ∀n∈S,P(n)\forall n\in S,\quad P(n)

    Even integer

    n=2k,k∈Zn=2k,\quad k\in\mathbb Z

    Odd integer

    n=2k+1,k∈Zn=2k+1,\quad k\in\mathbb Z

    Divisibility

    d∣n⟺n=dk for some k∈Zd\mid n\quad\Longleftrightarrow\quad n=dk\text{ for some }k\in\mathbb Z

    Consecutive integers

    n, n+1, n+2,…(n∈Z)n,\ n+1,\ n+2,\ldots\quad(n\in\mathbb Z)

    Direct-proof chain

    P ⟹ R1 ⟹ R2 ⟹⋯⟹ QP\ \Longrightarrow\ R_1\ \Longrightarrow\ R_2\ \Longrightarrow\cdots\Longrightarrow\ Q

    Identity notation

    f(x)≡g(x)f(x)\equiv g(x)

    Difference of squares

    A2−B2=(A−B)(A+B)A^2-B^2=(A-B)(A+B)

    Square is non-negative

    u2≥0for every u∈Ru^2\ge0\quad\text{for every }u\in\mathbb R

    Even target form

    E=2K,K∈ZE=2K,\quad K\in\mathbb Z

    Odd target form

    O=2K+1,K∈ZO=2K+1,\quad K\in\mathbb Z

    Odd-square result

    n odd⟹n2≡1(mod8)n\text{ odd}\quad\Longrightarrow\quad n^2\equiv1\pmod8

    Divisibility target

    d∣N⟺N=dK for some K∈Zd\mid N\quad\Longleftrightarrow\quad N=dK\text{ for some }K\in\mathbb Z

    Linear-combination property

    d∣a ∧ d∣b⟹d∣(ra+sb),r,s∈Zd\mid a\ \land\ d\mid b\quad\Longrightarrow\quad d\mid(ra+sb),\quad r,s\in\mathbb Z

    Consecutive integers

    n, n+1, n+2n,\ n+1,\ n+2

    Consecutive even integers

    2k, 2k+2, 2k+42k,\ 2k+2,\ 2k+4

    Consecutive odd integers

    2k+1, 2k+3, 2k+52k+1,\ 2k+3,\ 2k+5

    Counterexample test

    P(c) true and Q(c) false⟹∀x [P(x)⇒Q(x)] is falseP(c)\text{ true and }Q(c)\text{ false}\quad\Longrightarrow\quad \forall x\,[P(x)\Rightarrow Q(x)]\text{ is false}
    Topic 1.7Not started

    Rational exponents and logarithm laws

    Connect roots with fractional powers, apply exponent and logarithm laws accurately, and solve exponential or logarithmic equations while respecting domain restrictions.

    125 min estimated

    Topic 1.7 progress0%
    Learning objectives
    • Interpret rational exponents and convert fluently between radical and exponent notation.
    • Apply exponent laws to rational powers and simplify exact expressions.
    • Use logarithm definitions, identities, and domain restrictions correctly.
    • Expand and condense logarithmic expressions using product, quotient, and power laws.
    • Use change of base and GDC methods to evaluate logarithms.
    • Solve exponential and logarithmic equations and reject invalid solutions.
    Key formulas

    Rational exponent definition

    amn=amn=(an)ma^{\frac{m}{n}}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m

    Negative rational exponent

    a−mn=1amna^{\frac{-m}{n}}=\frac{1}{a^{\frac{m}{n}}}

    Special root form

    a1n=ana^{\frac{1}{n}}=\sqrt[n]{a}

    Radical–exponent conversion

    xmn=xmn,x>0, m∈Z, n∈Z>0\sqrt[n]{x^m}=x^{\frac{m}{n}},\qquad x>0,\ m\in\mathbb Z,\ n\in\mathbb Z_{>0}

    Principal square root

    x2=∣x∣\sqrt{x^2}=|x|

    Odd root of a power

    x2k+12k+1=x\sqrt[2k+1]{x^{2k+1}}=x

    Product law

    aras=ar+sa^r a^s=a^{r+s}

    Quotient law

    aras=ar−s\frac{a^r}{a^s}=a^{r-s}

    Power of a power

    (ar)s=ars(a^r)^s=a^{rs}

    Power of a product

    (ab)r=arbr(ab)^r=a^r b^r

    Power of a quotient

    (ab)r=arbr\left(\frac ab\right)^r=\frac{a^r}{b^r}

    Negative exponent conversion

    x−r=1xrx^{-r}=\frac{1}{x^r}

    Common-denominator strategy

    xpqxrs=x(ps+rq)(qs)x^{\frac{p}{q}}x^{\frac{r}{s}}=x^{\frac{(ps+rq)}{(qs)}}

    Exact before decimal

    amn=amn\sqrt[n]{a^m}=a^{\frac{m}{n}}

    Logarithm definition

    log⁡ax=y⟺ay=x\log_a x=y\quad\Longleftrightarrow\quad a^y=x

    Logarithm of one

    log⁡a1=0\log_a1=0

    Logarithm of its base

    log⁡aa=1\log_a a=1

    Inverse relationships

    alog⁡ax=x,log⁡a(ax)=xa^{\log_a x}=x,\qquad \log_a(a^x)=x

    Product law

    log⁡a(xy)=log⁡ax+log⁡ay\log_a(xy)=\log_a x+\log_a y

    Quotient law

    log⁡a(xy)=log⁡ax−log⁡ay\log_a\left(\frac{x}{y}\right)=\log_a x-\log_a y

    Power law

    log⁡a(xr)=rlog⁡ax\log_a(x^r)=r\log_a x

    Reciprocal result

    log⁡a(1x)=−log⁡ax\log_a\left(\frac1x\right)=-\log_a x

    Change of base

    log⁡ax=log⁡bxlog⁡ba\log_a x=\frac{\log_b x}{\log_b a}

    Calculator forms

    log⁡ax=ln⁡xln⁡a=log⁡xlog⁡a\log_a x=\frac{\ln x}{\ln a}=\frac{\log x}{\log a}

    Condensing pattern

    rlog⁡ax−slog⁡ay=log⁡a(xrys)r\log_a x-s\log_a y=\log_a\left(\frac{x^r}{y^s}\right)

    Same-base principle

    au=av⟹u=va^u=a^v\quad\Longrightarrow\quad u=v

    Logarithmic solution step

    af(x)=k⟹f(x)=ln⁡kln⁡aa^{f(x)}=k\quad\Longrightarrow\quad f(x)=\frac{\ln k}{\ln a}

    Log equality principle

    log⁡au=log⁡av⟹u=v\log_a u=\log_a v\quad\Longrightarrow\quad u=v

    Fractional exponent

    amn=amna^{\frac{m}{n}}=\sqrt[n]{a^m}

    Three log laws

    log⁡a(xy)=log⁡ax+log⁡ay,log⁡a(xy)=log⁡ax−log⁡ay,log⁡a(xr)=rlog⁡ax\log_a(xy)=\log_ax+\log_ay,\quad \log_a(\frac{x}{y})=\log_ax-\log_ay,\quad \log_a(x^r)=r\log_ax

    Change of base

    log⁡ax=ln⁡xln⁡a\log_a x=\frac{\ln x}{\ln a}
    Topic 1.8Not started

    Infinite geometric series

    Understand an endless geometric sum through its partial sums, test whether it converges, calculate its limiting value, and apply the model to algebraic and real-world problems.

    110 min estimated

    Topic 1.8 progress0%
    Learning objectives
    • Distinguish an infinite geometric sequence from an infinite geometric series.
    • Identify the first term and common ratio of a geometric series.
    • Determine convergence using the condition ∣r∣<1|r|<1.
    • Derive and apply the sum-to-infinity formula.
    • Find unknown terms, first terms, or common ratios from series information.
    • Explain why a series diverges when the convergence condition fails.
    • Model recurring decimals, repeated doses, and bouncing-ball distances.
    • Use a GDC to investigate partial sums and solve unknown-ratio problems.
    Key formulas

    Infinite geometric series

    a+ar+ar2+ar3+⋯a+ar+ar^2+ar^3+\cdots

    Partial sum notation

    Sn=∑k=0n−1arkS_n=\sum\limits_{k=0}^{n-1}ar^k

    Common ratio

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

    Geometric term

    un=arn−1u_n=ar^{n-1}

    Geometric convergence condition

    ∣r∣<1⟺−1<r<1|r|<1\quad\Longleftrightarrow\quad -1<r<1

    Vanishing powers

    ∣r∣<1⟹rn→0|r|<1\quad\Longrightarrow\quad r^n\to0

    Finite geometric sum

    Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r}

    Sum to infinity

    S∞=a1−r,∣r∣<1S_\infty=\frac{a}{1-r},\qquad |r|<1

    First term from the sum

    a=S∞(1−r)a=S_\infty(1-r)

    Ratio from first term and sum

    r=1−aS∞,S∞≠0r=1-\frac{a}{S_\infty},\qquad S_\infty\ne0

    Term condition

    un=arn−1u_n=ar^{n-1}

    Remainder after n terms

    Rn=S∞−Sn=arn1−rR_n=S_\infty-S_n=\frac{ar^n}{1-r}

    Approximation relation

    S∞=Sn+RnS_\infty=S_n+R_n

    Bouncing-ball distance

    D=h+2hq1−q=h(1+q1−q)D=h+\frac{2hq}{1-q}=h\left(\frac{1+q}{1-q}\right)

    Repeated fixed contribution

    A∞=d1−r,∣r∣<1A_\infty=\frac{d}{1-r},\qquad |r|<1

    Two-digit recurring decimal

    0.ab‾=10a+b99,a,b∈{0,1,…,9}0.\overline{ab}=\frac{10a+b}{99},\qquad a,b\in\{0,1,\ldots,9\}

    Partial-sum calculator entry

    Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r}

    Solver interval

    −1<r<1-1<r<1

    Convergence and sum

    ∣r∣<1⟹S∞=a1−r|r|<1\quad\Longrightarrow\quad S_\infty=\frac{a}{1-r}

    Term and remainder

    un=arn−1,Rn=arn1−ru_n=ar^{n-1},\qquad R_n=\frac{ar^n}{1-r}
    Topic 1.9Not started

    Binomial theorem

    Use Pascal’s triangle and combination notation to expand integer powers of binomials, identify individual terms and coefficients, and solve constant-term and parameter problems efficiently.

    115 min estimated

    Topic 1.9 progress0%
    Learning objectives
    • Explain the structure of the binomial theorem for non-negative integer powers.
    • Generate and interpret rows of Pascal’s triangle.
    • Calculate binomial coefficients using combination notation and a GDC.
    • Expand binomials containing variables, negative terms, and fractional coefficients.
    • Use the general term to find a specified term or coefficient without a full expansion.
    • Find constant terms by solving for the required variable exponent.
    • Use coefficient identities and substitutions to solve IB-style problems.
    Key formulas

    Binomial theorem

    (a+b)n=∑r=0n(nr)an−rbr(a+b)^n=\sum\limits_{r=0}^{n}\binom nr a^{n-r}b^r

    Number of formal binomial summands

    formal summands before collection=n+1\text{formal summands before collection}=n+1

    Pascal recurrence

    (nr)=(n−1r−1)+(n−1r)\binom nr=\binom{n-1}{r-1}+\binom{n-1}{r}

    Coefficient symmetry

    (nr)=(nn−r)\binom nr=\binom n{n-r}

    Row sum

    ∑r=0n(nr)=2n\sum\limits_{r=0}^{n}\binom nr=2^n

    Combination formula

    (nr)=n!r!(n−r)!\binom nr=\frac{n!}{r!(n-r)!}

    Factorial

    n!=n(n−1)(n−2)⋯2⋅1,0!=1n!=n(n-1)(n-2)\cdots2\cdot1,\qquad 0!=1

    Boundary values

    (n0)=(nn)=1\binom n0=\binom nn=1

    Expanded form

    (a+b)n=(n0)an+(n1)an−1b+⋯+(nn)bn(a+b)^n=\binom n0a^n+\binom n1a^{n-1}b+\cdots+\binom nnb^n

    Exponent check

    (n−r)+r=n(n-r)+r=n

    General term

    Tr+1=(nr)an−rbrT_{r+1}=\binom nr a^{n-r}b^r

    pth-term index

    r=p−1r=p-1

    Coefficient method

    power of x=k⟹solve for r\text{power of }x=k\quad\Longrightarrow\quad \text{solve for }r

    Quadratic coefficient

    (n2)=n(n−1)2\binom n2=\frac{n(n-1)}2

    Constant-term condition

    net exponent of x=0\text{net exponent of }x=0

    Admissible index

    r∈{0,1,2,…,n}r\in\{0,1,2,\ldots,n\}

    Sum of coefficients

    ∑k=0nck=P(1)\sum\limits_{k=0}^{n}c_k=P(1)

    Alternating coefficient sum

    ∑k=0n(−1)kck=P(−1)\sum\limits_{k=0}^{n}(-1)^kc_k=P(-1)

    GDC term calculation

    (nr)an−rbr\binom nr a^{n-r}b^r

    Core theorem

    (a+b)n=∑r=0n(nr)an−rbr(a+b)^n=\sum\limits_{r=0}^{n}\binom nr a^{n-r}b^r

    Target term

    Tr+1=(nr)an−rbrT_{r+1}=\binom nr a^{n-r}b^r

    Constant condition

    net power of x=0\text{net power of }x=0
  2. Unit 2
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    Functions

    Functions are one of the central threads of AA SL. Students learn to represent relationships, transform graphs, solve equations and connect symbolic structure with graphical behavior.

    11 topics

    Unit 2 progress0%
  3. Unit 3
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    Geometry and Trigonometry

    This topic combines spatial reasoning, coordinate geometry and trigonometric functions. It moves from triangle methods to radians, identities, periodic functions and trigonometric equations.

    8 topics

    Unit 3 progress0%
  4. Unit 4
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    Statistics and Probability

    Statistics and probability develop the ability to analyze data, quantify uncertainty and use probability distributions. Technology is important, but interpretation and model choice remain central.

    12 topics

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  5. Unit 5
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    Calculus

    Calculus is a major emphasis of AA SL. The topic connects rates of change, graph behavior, optimization, motion and accumulation through differentiation and integration.

    11 topics

    Unit 5 progress0%