Mathematics · May / June 2021 · Paper 21 · Question 3
YMA01
As printed in the paper
Show the answer and mark scheme
3(i)[2]
2 +
3
2
8
2
10
3 +
3
3
27
3
30
For p = 2 :
= +
=
; For p = 3 :
=
+
=
For p = 5 :
5 +
3
5
=
125
+
5
=
130
; For p = 7 :
7 +
3
7
=
343
+
7
=
350
Each case gives a multiple of 10. As 2,3,5 and 7 are the only single digit
primes, the result has been proved for all single digit primes.
(
)
3
3(ii)[3]
n
+
1
−
n
3
=
n
3
+
3
n
2
+
3
n
+ −
1
n
3
=
3
n
2
+
3
n
+
1
3(
2
)
1
which is one more than a multiple of 3, so is not divisible by
=
n
+
n
+
3 for any n∈∞
Notes:
(i)
M1: Checks result for at least three of the four single digit primes (2,3,5 and 7) – attaining a
multiple of 10 is enough, no need to see the product. Allow if there are slips.
A1: All four cases correctly checked, with minimal conclusion that the result is true. Ignore checks
3
on nonprime values such as p = 8 which gives
8
+
8
=
520
, but award A0 if the case p = 1, leads to
a conclusion that the result is not true.
(ii)
M1: Expands to a four term cubic (may have incorrect coefficients) and then cancels the n3 terms
(if n = 2k and n= 2k + 1 are used, both must be attempted.)
A1: Correct quadratic in n. (Both correct if n = 2k and n= 2k + 1 are used)
A1: Correct explanation and conclusion given. For the explanation accept e.g factors out 3 from the
(
)
2
relevant terms to achieve a form
3
×
n
+
n
+ which is one more than a multiple of 3, or explains
1
each term other than 1 is divisible by 3
13
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