Free SACE Specialist Mathematics Flashcard Review
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What is P(n) in an induction proof of 1+3+⋯+(2n−1)=n²?
P(n) is the complete statement “the sum of the first n positive odd integers equals n²”, defined for each positive integer n.
Error check: can a proof of P(n) for all n≥1 begin only at n=2?
No. The required first case P(1) remains unproved; verify 1=1² before the inductive implication.
In an induction step, exactly what may be assumed about k?
Assume P(k) is true for one arbitrary positive integer k; do not assume P(k+1), which is what must be derived.
Application: if S_k=k² for the first k odd numbers, compute S_(k+1).
The next term is 2k+1, so S_(k+1)=k²+(2k+1)=(k+1)².
Which sentence completes a proof after base and successor cases?
Because P(1) holds and P(k) implies P(k+1) for every k≥1, P(n) holds for every positive integer n by mathematical induction.
Application: if 7^k−1 is divisible by6, rewrite 7^(k+1)−1.
7^(k+1)−1=7(7^k−1)+6; both terms are multiples of6 under the hypothesis.
Error check: proving a series formula at k, which term is added for n=k+1?
Add the term indexed k+1. Re-adding the k-th term repeats an old summand and breaks the successor calculation.
How does induction handle a product ∏_(r=1)^n a_r?
Assume the product formula through k, multiply by a_(k+1), then simplify to the asserted k+1 formula; check the first product separately.
Application: if h^(k)(x)=e^x(x+k), find h^(k+1)(x).
Differentiate by the product rule: e^x(x+k)+e^x=e^x(x+k+1).
Application: advance (cis θ)^k=cis(kθ) by one power.
Multiply by cis θ and use the compound-angle identities: cis(kθ)cis θ=cis((k+1)θ).
What is the key step for the extended triangle inequality by induction?
From |z₁+⋯+z_k|≤Σ_(r=1)^k|z_r|, apply |u+v|≤|u|+|v| to add z_(k+1).
Scope check: should an induction drill introduce unrelated recursion or advanced inequalities?
No. Keep the proposition within the Stage 2 induction focus, such as a sum, divisibility, product, derivative pattern or de Moivre identity.
Application: multiply (2+i)(1−3i).
Distribute and use i²=−1: 2−6i+i−3i²=5−5i.
For z=−3+4i, identify Re z and Im z.
Re z=−3 and Im z=4; the imaginary part is the real coefficient 4, not the term 4i.
Application: convert −1+i√3 to modulus-argument form.
The modulus is2 and the point is in quadrant II with argument2π/3, so z=2cis(2π/3).
How do modulus and argument behave in a nonzero product zw?
|zw|=|z||w| and arg(zw)=arg z+arg w modulo2π. Addition of moduli is incorrect.
What geometric action does multiplication by 2cis(π/3) perform?
It doubles distance from the origin and rotates anticlockwise by π/3.
Application: divide 6cis(5π/6) by 2cis(π/3).
Divide moduli to3 and subtract arguments: 5π/6−π/3=π/2, giving 3i.
Application: compute [cis(π/7)]^7.
By de Moivre, [cis(π/7)]^7=cisπ=−1.
Error check: is z^(1/2)=r^(1/2)cis(θ/2) the only square root?
No. A nonzero complex number has two square roots, with arguments θ/2 and θ/2+π; negative powers also require z≠0.