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} , n∈Z+ , satisfies the recurrence relation un+2=5un+1−6un .
Given that u1=u2=3 , obtain an expression for un in terms of n .
The sequence {vn} , n∈Z+ , satisfies the recurrence relation vn+2=4vn+1−4vn .
Given that v1=2 and v2=12 , use the principle of strong mathematical induction to show that vn=2n(2n−1) .
Markscheme
the auxiliary equation is
m2−5m+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=3×2n−3n A1
[6 marks]
we first prove that vn=2n(2n−1) 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]