Question 2:
2 | (a) | 2u – 7u + 3u = 0 & u = α rn
n+2 n+1 n n
⇒ 2r2 – 7r + 3 = 0
⇒ r = 3 or r = 1
2
So GS is (u =)A×3n +B (1)n
n 2 | M1
A1
[2] | 1.1
1.1 | Deriving the auxiliary equation.
3 term quadratic equation. “= 0”
could be implied by correct solution.
B 1
or A×3n + or A×3n +B×
2n 2n
or A×3n +B×2 −n or A×3n +B×0.5n | If M0 then SCB1 for correct answer.
Condone missing brackets around the
fraction.
If deduced from repeated substitution
then must be fully correct for M1A1.
2 | (b) | Trial function: Try u = an2 + bn + c
n
2[a(n + 2)2 + b(n + 2) + c] –
7[a(n + 1)2 + b(n + 1) + c] + 3(an2 + bn + c)
= 2[an2 + (4a + b)n + 4a + 2b + c] –
7[an2 + (2a + b)n + a + b + c] + 3(an2 + bn + c)
= (2a – 7a + 3a)n2 + (8a + 2b – 14a – 7b + 3b)n +
(8a + 4b + 2c – 7a – 7b – 7c + 3c)
= –2an2 + (–6a – 2b)n + (a – 3b – 2c)
= 20n2 + 60n
⇒ –2a = 20, –6a – 2b = 60 and a – 3b – 2c = 0
⇒ a = –10, b = 0, c = –5
So GS is u = A×3n +B (1)n −10n2−5
n 2
, | B1
M1
M1
A1
B1FT
[5] | 3.1a
1.1
1.1
1.1
1.1 | Correct general form for trial
function. Either stated as u or as TF
n
or clearly subbed in.
Substituting their polynomial form
correctly into LHS of recurrence
relation.
Expanding, collecting terms and
comparing coefficients with correct
RHS coefficients to derive three
equations in a, b and c (may be
progressively solved).
FT their CF (GS from (a)) + PTF
provided that CF has two arbitrary
constants. Only ISW if this is
labelled as GS. | Other terms eg dn3 may be added; B1
can be awarded when eg d = 0 stated
(from working or assumption)
May also see CF subbed in.
u – 7u + 3u = 20n2 – 20n – 40
n n – 1 n – 2
M1 can be earned after single minor
slip in indexing.
M1 can be earned from their TF
provided it does not lead to significant
reduction in complexity.
Must be “u =”
n
Do not accept missing brackets in
fraction for B1FT.
2 | (c) | n = 0 ⇒ –9 = A + B – 5
n=1⇒−12=3A+1B−10−5
2
So 6A + (–4 – A) = 6
⇒ A = 2, B = –6 so PS is
u =2×3n−6 (1)n −10n2−5 cao
n 2 | M1
M1
A1
[3] | 1.1
1.1
1.1 | Using one initial condition in their
GS from (b) to derive an equation in
A and B.
Using another initial condition in
their GS from (b) to derive a second
equation in A and B and attempt to
solve (can be implied by correct
solution).
Could be BC. | A + B = –4
6A + B = 6
M0M0A0 for using just the CF.
A + B = –9, 3A + ½B = –12,
A = –3, B = –6
“u =” and fraction in brackets but only
n
penalise either once in 2(b) and 2(c).
2 | (d) | u = 2×9 – 6/4 – 10×4 – 5 (= –28.5)
2
u = 2×27 – 6/8 – 10×9 – 5 (= –41.75)
3
u = 2×81 – 6/16 – 10×16 – 5 (= –3.375)
4
57 167 27
NB −28.5=− , −41.75=− , −3.375=−
2 4 8
(So u , u , u , ... is an increasing sequence so the
3 4 5
least value is u =) –41.75
3 | M1
A1 | 3.1a
2.2a | Using their PS (of the form
n
 1
A×3n+B +αn2+βn+γ) to
 
2
calculate any three consecutive terms
u , n ≥ 2 (eg u , u and u ).
n 2 3 4
Values for u , u and u all correct
2 3 4
and –41.75 explicitly selected.
Not just “n = 3” as final answer.
Correct value unsupported is 0/2. | Values do not have to be correct
provided that it is clear what is being
attempted.
M0 if it is clear that the RR is being
used to generate values.
If M0 then SC1 for correct values
found from the RR and correct
conclusion
Alternative method: | M1 | Using their PS to calculate, u – u
3 2
and u – u , perhaps from derived
4 3
general rule. Values do not have to
be correct provided that it is clear
what is being attempted.
u −u =2×3n+1−6 (1)n+1 −10(n+1)2−5
n+1 n 2
( )
− 2×3n −6 (1)n −10n2−5
2
=6×3n −2×3n −3 (1)n +6 (1)n −20n−10
2 2
=4×3n +3 (1)n −20n−10
2
n=2:u −u =−13.25 (<0)
3 2
n=3:u −u =38.375 (>0)
4 3
So u , u , u , ... is an increasing sequence so the
3 4 5 | All values correct.
least value is (u =) –41.75
3 | A1 | Correct value unsupported is 0/2.
[2]
M1
Using their PS to calculate, u – u
3 2
and u – u , perhaps from derived
4 3
general rule. Values do not have to
be correct provided that it is clear
what is being attempted.