Question 2:
2 | (a) | 4 t t 0 t 4 n  − =  =  − oe
n 1 n n +
Try t = p n + q
n
4(p(n + 1) + q) – (pn + q) = 15n + 17
(3p = 15 and 4p + 3q = 17 =>) p = 5, q = –1 oe
So GS is ( t ) 5 n 1 4 n  = − +  −
n

t =5n−1+4 −n and t =22=4+
n 1
4
so  = ...
... –8 so t = 5 n − 1 − 8  4 − n oe (eg
n
t =5n−1−24 −n+1 )
n | B1
M1
M1
A1
B1FT
M1
A1
[7] | 1.1
1.1
1.1
1.1
1.1
1.1
1.1 | Correct form for complementary
function (must include arbitrary
constant). “+ ” is B0 unless
legitimately recovered.
Correct form for trial function
(could be higher polynomial etc if
other coefficients found to be 0).
Substituting their form correctly
into correct recurrence relation
Their numerical PTF plus their
CF with one arbitrary constant
originally.
Using initial condition t = 2
1
correctly in their GS (= PTF + CF
with one arb const) to find 
Full form for solution must be
seen (including “t =”) but ISW.
n | n
1  
Could see eg 0.25n or 
4

or or 4 −(n+1) etc
4 n
Must have “t =...” or “trial
n
function is...” oe or see clear
evidence of substitution into RR.
4t – t = 15n + 2 =>
n n–1
4(pn + q) – (p(n–1) + q) = 15n +
2
Condone missing brackets for
(¼)n for B1 if used correctly in
next step.
Substituting into just CF is M0
n
1
Do not condone for A1.
4
2 | (b) | (i) | Because it is non-linear | B1
[1] | 2.4 | or because the coefficients are not
constant or because it cannot be
written in the form
u = c u + f ( n ) where c is a
n + 1 n
constant oe | Must answer the question or be
readable as an answer to the
question.
Ignore other comments unless
egregiously incorrect or directly
contradictory
2 | (b) | (ii) | n = 1  a + b = 2
( n + 1 ) u − u 2n
n + 1
2
 b   b 
= ( n + 1 ) a ( n + 1 ) + − a n +
n + 1 n
b2
=an2+2an+a+b−a2n2−2ab−
n2
b2
=(a−a2)n2+2an+a+b−2ab−
n2
1
=2n−
n2
eg n term  a = 1 and const term  b = 1
hence b2 = 1(so n–2 term is correct) and
coefficient of n2 is zero, as required | B1
*M1
dep*M1
A1
A1 | 1.1
1.1
1.1
1.1
2.2a | Substituting n = 1 into the given
formula for u .
n
Correctly substituting for both
u and u in (LHS of) RR
n+1 n
Simplifying LHS correctly with
n2 and constant terms collected
(can be implied by correct
equations: 2a = 2, a – a2 = 0, a +
b – 2ab = 0, (–)b2 = (–)1)
(Equating to RHS and)
comparing coefficients twice, or
once and using a + b = 2, to
derive values of a and b www.
Verifying properly that all the
other coefficients equate/cancel
and that u = 2.
1
Could also be done by
substituting solution with a = b =
1 back into the recurrence
relation. u = 2 must be properly
1
established (but after B1 this
could be done by stating a + b =
2). | This could be done as verification
that u = 2 at the end, after
1
a = b = 1 has been established.
If b = 2 – a used then:
 2 − a   2 − a 
( n + 1 ) a ( n + 1 ) + − a n +
n + 1 n
=an2+2an+a+2−a−a2n2
(2−a)2
−4a+2a2−
n2
=(a−a2)n2+2an+2
(2−a)2
−4a+2a2−
n2
If M1M0 then SCB1 for a = 1,
b = 1 found without justification
Could see eg a + b = 2 and a + b
– 2ab = 0 solved simultaneously
IF B0M1M1A1 then SCA1 can
be awarded without establishing
u = 2.
1
Alternative method:
n = 1  a + b = 2 | B1 | 1.1
Substituting n = 1 into the given
formula for u .
n
12
n = 1 , u = 2  2 u − 2 u = 2  1 − = 1
1 2 1
1
5
 2 u = 1 + 2 2 = 5  u =
2 2
2 | *B1 | 1.1 | Substituting n = 1 and u = 2 into
1
the RR correctly to derive u
2
b 5
 2 a + =
2 24
 4 a + = b a + 2 − a = 3 a + 2 = 5
 a = 1 , b = 1 | dep*
M1* | 1.1 | Using the suggested solution on
u to derive a second equation in
2
a and b and solving
simultaneously | Award for a = b = 1 conjecture
based on u = 2/1, u = 5/2, u =
1 2 3
10/3, u = 17/4, ..., u = (n2 + 1)/n
4 n
L H S = ( n + 1 ) u − u 2n
n + 1
2
 1   1 
= ( n + 1 ) n + 1 + − n +
n + 1 n | dep**
M1 | 1.1 | Correctly substituting for u and
n+1
u in (LHS of) RR with their
n
values for a and b
1
=n2 +2n+1+1−n2 −2−
n2
1
=2n− =RHS as required
n2 | 1
=n2 +2n+1+1−n2 −2−
n2
1
=2n− =RHS as required
n2 | A1 | 1.1 | Simplifying LHS and, after
opening all brackets, showing it
equals given RHS with a = b = 1.
Some observation or conclusion
is required
[5]
Alternative method:
n = 1  a + b = 2
B1
1.1
Substituting n = 1 and u = 2 into
1
the RR correctly to derive u
2
Using the suggested solution on
u to derive a second equation in
2
a and b and solving
simultaneously
Award for a = b = 1 conjecture
based on u = 2/1, u = 5/2, u =
1 2 3
10/3, u = 17/4, ..., u = (n2 + 1)/n
4 n
dep*
M1*
Correctly substituting for u and
n+1
u in (LHS of) RR with their
n
values for a and b
dep**
M1
[2]