Standard +0.3 This is a straightforward algebraic proof requiring students to represent consecutive odd numbers as (2n+1) and (2n+3), expand their cubes, and show the sum simplifies to a multiple of 4. While it requires careful algebraic manipulation across several steps, the approach is standard and the technique of representing odd numbers algebraically is well-practiced at AS level, making it slightly easier than average.
Question 8:
8 | Expresses two consecutive odd
numbers as
(2k ± 1)
or (2k + 1) & (2k + 3)
or (k ± 1) where k is even
or (k + 1) & (k + 3) where k is
even
OE | 2.1 | M1 | Let the consecutive odd numbers
be (2k + 1) and (2k – 1) where k is
an integer
(2k + 1)3 = 8k3 + 12k2 + 6k + 1
(2k – 1)3 = 8k3 – 12k2 + 6k – 1
Sum = 16k3 + 12k
= 4k(4k2 +3)
Factor of 4 shows that this is a
multiple of 4
Expands at least one odd-
numbered cubic expression
– allow one slip | 1.1a | M1
Expands both of their two
different odd-numbered cubic
expressions correctly | 1.1b | A1
Simplifies the sum of their cubic
expansions correctly | 1.1a | M1
Identifies common factor of 4
and completes proof.
If (k ± 1) or (k + 1) & (k + 3)
have been used, reference must
be made to k being even, to
clearly identify the factor of 4.
CSO
N.B.
(2k + 3)3 = 8k3 + 36k2 + 54k + 27
(k – 1)3 = k3 − 3k2 + 3k – 1
(k + 1)3 = k3 + 3k2 + 3k + 1
(k − 3)3 = k3 − 9k2 + 27k − 27
(2k + 1)3 + (2k + 3)3 =
16k3 + 48k2 + 60k + 28 | 2.1 | R1
Question 8 Total | 5
Q | Marking instructions | AO | Marks | Typical solution