IB Maths AA SL AA 1.3 Proof questions

Exam-style questions in the shape of Paper 1 and Paper 2, with full markschemes. Read the markscheme, or type, photograph or draw your answer and get it marked.

Question 1

Paper 1, no calculatorHard6 marks

For n∈Zn\in\mathbb Z define

Sn=(n+1)5+(n−1)5.S_n=(n+1)^5+(n-1)^5.

A student conjectures that SnS_n is divisible by 3232 for every odd integer nn

(a)

Calculate S1S_1 directly.

[1]
(b)

Starting from SnS_n prove the identity

Sn≡2n5+20n3+10nS_n\equiv 2n^5+20n^3+10n

for all n∈Zn\in\mathbb Z Use == between successive expressions and ≡\equiv for the final identity.

[2]
(c)

Hence prove the student’s conjecture for every odd integer nn

[3]

Question 2

Paper 1, no calculatorHard6 marks

Let m∈Z+m\in\mathbb Z^+ Figure 1 shows a rectangle with side lengths m+1m+1 and m+2m+2 divided into four smaller rectangles.

It is claimed that integers AA and BB can be chosen so that, for every m∈Z+m\in\mathbb Z^+

1m+1+1m+2+Am+Bm2+3m+2=3m+1.\frac{1}{m+1}+\frac{1}{m+2}+\frac{Am+B}{m^2+3m+2} =\frac{3}{m+1}.
2m2m2mm+1m12mm+2
(a)

Using the areas shown in Figure 1, write an identity relating the area of the whole rectangle to the areas of the four smaller rectangles. Use ≡\equiv to denote an identity.

[1]
(b)

If the claim is true, the equality for m=1m=1 requires a particular relation between AA and BB Determine this relation.

[1]
(c)

Determine AA and BB so that the claim is an identity for every m∈Z+m\in\mathbb Z^+ Then, starting with the left-hand side, prove the resulting identity. Use your answer to part (b).

[4]