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Date None Specimen Marks available 6 Reference code SPNone.3dm.hl.TZ0.5
Level HL only Paper Paper 3 Discrete mathematics Time zone TZ0
Command term Obtain Question number 5 Adapted from N/A

Question

The sequence {un} , nZ+ , satisfies the recurrence relation un+2=5un+16un .

Given that u1=u2=3 , obtain an expression for un in terms of n .

[6]
a.

The sequence {vn} , nZ+ , satisfies the recurrence relation vn+2=4vn+14vn .

Given that v1=2 and v2=12 , use the principle of strong mathematical induction to show that vn=2n(2n1) .

[8]
b.

Markscheme

the auxiliary equation is

m25m+6=0     M1

giving m=2, 3     A1

the general solution is

un=A×2n+B×3n     A1

substituting n = 1, 2

2A+3B=3     M1

4A+9B=3     A1

the solution is A = 3, B = –1 giving un=3×2n3n     A1

[6 marks]

a.

we first prove that vn=2n(2n1) for n = 1, 2     M1

for n = 1, it gives 2×1=2 which is correct

for n = 2 , it gives 4×3=12 which is correct     A1

we now assume that the result is true for n     M1

consider

{v_{k + 1}} = 4{v_k} - 4{v_{k - 1}}{\text{ }}(k \geqslant 2)     M1

= {4.2^k}(2k - 1) - {4.2^{k - 1}}(2k - 3)     A1

= {2^{k + 1}}(4k - 2 - 2k + 3)     A1

= {2^{k + 1}}\left( {2(k + 1) - 1} \right)     A1

this proves that if the result is true for n \leqslant k then it is true for n \leqslant k + 1

since we have also proved it true for n \leqslant 2 , the general result is proved by induction     R1

Note: A reasonable attempt has to be made to the induction step for the final R1 to be awarded.

 

[8 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 10 - Option: Discrete mathematics » 10.11 » Recurrence relations. Initial conditions, recursive definition of a sequence.

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