Mathematics · Oct / Nov 2024 · Paper 21 · Question 2
YMA01
As printed in the paper
The question as text
2. A sequence of numbers u1, u2, u3, ... is defined by u1 = 7 n un + k un+1 = (–1) where k is a constant.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
2(a)[3]
=
7
=
−
7
u
u
k
1
2
(
)
u
=
k
−
7
u
=
k
+
k
−
7
u
=
k
−
2
k
−
7
2
3
4
=
+ −+
(
7
)
=
7*
u
k
k
5
2(b)[2]
4
=
30
7
+
−+
7
2
−−
7
+
7
=
30
=
...
u
k
k
k
k
r
=
1
r
k =
15
2(c)[2]
150
u
=
37 30
×
+
7
+
"15" 7
−
r
r
=
1
=
1125
General Guidance
Note that it is possible to obtain correct answers in all parts of this question fortuitously.
The A marks should only be awarded following correct work.
(a)
B1:
Correct expression for
2u e.g.
k −
7
with no incorrect work seen e.g.
(
)
= −
1
7
×+
7
= −+
7
scores B0. Must be seen in part (a).
u
k
k
2
Allow e.g.
(
)
1
or e.g.
(
1
)
7
or e.g.
u
= −
1
×+
7
k
u
= −
×+
k
u
= −×+
1 7
2
2
2
M1:
Attempts to use the recurrence formula to reach the 5th term but condone consistent
use of (n + 1) for n on the (
1
)
n
term so e.g.
(
1
)
2
7
is condoned but e.g.
−
u
= −
×+
k
2
(
)
7
u
= −
1
×+
7
k
is M0
2
Allow even if any slips mean their sequence is not of order 4.
A1*: Obtains
u =
7
with no errors or omissions. The
"
u = must appear at some point but
"
5
5
allow e.g.
"
u + =
"
4 1
For reference the terms are:
u
=
7,
u
=
k
−
7,
u
=
2
k
−
7,
u
=
7
−
k u
,
=
1
2
3
4
5
(a) Alternatives:
1
= −
+
,
= −
+
2 ,
=
−
,
=
=
7
u
u
k u
u
k u
u
k u
u
2
1
3
1
4
1
5
1
B1: Implied by
= −
+
u
u
k
2
1
M1: Attempts to use the recurrence formula to reach the 5th term but condone use of
(n + 1) for n on the (
)
n
−
1
term as above
A1*: Fully correct work to reach
u =
7
5
(Allow the above approach to be used with a made up k or a wrong k as the result is
independent of k)
2
u
=
7
u
=
7
−
k
u
=
2
k
−
7
u
=
k
−
7
u
=
7
Hence true.
5
4
3
2
1
B1:
u
=
7
−
k
4
M1: Attempts to use the recurrence formula correctly backwards to reach the 1st term but
condone use of (n + 1) for n on the (
1
)
n
term as above.
−
A1*: Fully correct work to reach
u =
7
with a (minimal) conclusion.
1
(b)
M1:
Adds their first 4 terms, sets = 30 and solves a linear equation for k
Condone slips in copying their 4 terms if the intention is clear.
Allow even if their sequence is of order 2 e.g. 7,
−
7, 7,
−
7, ...
k
k
A1:
Correct value from correct work and a correct sequence.
(c)
Marks in (c) are only available if their sequence is of order 4
M1:
Correct numerical attempt for the required sum using their k. May be implied by the
correct answer or the correct ft answer unless an incorrect method is seen e.g.
37 30
×
+
u
+
u
or
38 30
×
−
u
−
u
score M0
3
4
1
2
Condone 150
4 ×
30
or 75 15
×
or 37.5 30
×
following a correct sequence and the
correct value of k.
(
Allow any equivalent correct method e.g.
38 30
×
−
2 "15" 7
×
−+ −
7 "15"
A1:
For 1125 from correct work and a correct sequence.
Correct answer only scores both marks.
For reference the correct sequence is:
n
1
2
3
4
un
7
k − 7
2k − 7
7 – k
Value (with k = 15)
7
8
23
−8
An unusual but correct method in (c) is:
150
(
)
(
)
(
)
=
38 7
×+
38 15
−
7
+
37 30
−
7
+
37 7 15
−
=
1125
u
r
r
=
1
Example – Candidates who consistently use
(
)
+
1
u
+ = −
1
n
u
+
k
n
1
n
This gives:
(a)
=
7,
=
+
7,
= −
7,
=
−
7,
=
7
which scores B0M1A0
u
u
k
u
u
k
u
1
2
3
4
5
4
(b)
=
30
7
+
+
7
−+
7
−
7
=
30
=
15
which scores M1A0
u
k
k
k
r
=
1
r
150
(c)
u
=
37 30
×
+
7
+
"15" 7
+
=
1139
which scores M1A0
r
=
1
r
For reference this incorrect sequence is:
1
2
3
4
n
un
7
k + 7
−7
k – 7
Value (with k = 15)
7
22
8
−7
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