Question 8:
8 | Base case: when n=0,1170 −130 −1=9 which is
divisible by 3.
Assume that, when n=k,117n −13n −1 is divisible by
3.
That is, 1 1  7 k − 1 3 k − 1 = 3 m (for some integer m).
117k+1−13k+1−1
=7(3m+13k +1)−13k+1−1
= 3 ( 7 m + 2 − 2  1 3 k ) (which is divisible by 3).
So, if true for n = k this implies true for n = k + 1. True
when n = 0 so therefore true for all integers n 0 . | B1
M1*
M1dep*
A1
A1
[5] | 2.5
2.1
3.1a
2.2a
2.4 | Base case shown to be true – condone when n = 0, 11 – 1 – 1 = 9 or just 9,
must say that 9 is divisible by 3 or show explicitly e.g. 9 = 3  3 .
Allow base case with n = 1, 117−13−1=63 or just 63, must say that 63
is divisible by 3 or show explicitly e.g. 6 3 = 3  2 1 .
Forms correct inductive hypothesis as an equation in terms of two different
letters (oe in words e.g. 117k −13k −1is divisible by 3) – condone using
n e.g. 1 1  7 n − 1 3 n − 1 = 3 m for this mark.
Considers 117k+1−13k+1−1 and use inductive hypothesis that
1 1  7 k − 1 3 k − 1 = 3 m - must not be using n for k.
Shows using the inductive hypothesis that 1 1  7 k + 1 − 1 3 k + 1 − 1 is a multiple
of 3. Some common answers are 3 ( − 2 2  7 k + 4 + 1 3 m ) or
3(7m+2−213k) 3(m+227k −413k).
or
Conclusion, dependent on B1M1M1A1 and no errors seen in their proof.
Do not award this mark if n = 0 not used as base case. However, this mark
can be awarded if n = 1 used as base case and n = 0 considered separately.
Must mention, ‘n = k or P(k)’, ‘n = k + 1 or P(k + 1)’, ‘n = 0 or base case
or P(0)’ (provided P(n) is defined) and ‘n 0or all n oe’ condone ‘all
integers’ but not ‘positive integers’ or similar incorrect statement.
Alternative method
Base case: when n = 0 , 1 1  7 0 − 1 3 0 − 1 = 9 which is
divisible by 3. | Base case: when n = 0 , 1 1  7 0 − 1 3 0 − 1 = 9 which is | B1 | Base case shown to be true – condone when n = 0, 11 – 1 – 1 = 9 or just 9,
must say that 9 is divisible by 3 or show explicitly e.g. 9 = 3  3 .
Allow base case with n = 1, 117−13−1=63 or just 63, must say that 63
is divisible by 3 or show explicitly e.g. 6 3 = 3  2 1 .
B1 | Base case shown to be true – condone when n = 0, 11 – 1 – 1 = 9 or just 9,
divisible by 3.
must say that 9 is divisible by 3 or show explicitly e.g. 9 = 3  3 .
Allow base case with n = 1, 117−13−1=63 or just 63, must say that 63
Assume that, when n = k , 1 1  7 n − 1 3 n − 1 is divisible by
3.
That is, f ( k ) = 1 1  7 k − 1 3 k − 1 is divisible by 3. | M1* | Forms correct inductive hypothesis - condone using n e.g.
f ( n ) = 1 1  7 n − 1 3 n − 1 for this mark. Do not need to use f(k) – can just
state that 117k −13k −1 is divisible by 3.
f ( k + 1 ) − f ( k ) = ( 1 1  7 k + 1 − 1 3 k + 1 − 1 ) − ( 1 1  7 k − 1 3 k − 1 )
= 6  1 1  7 k − 1 2  1 3 k | M1dep* | Considers correct expression for f ( k + 1 ) − f ( k ) and simplifies to an
expression of the form  a  7 k  b  1 3 k - must not be using n for k.
f ( k + 1 ) = f ( k ) + 3 ( 2 2  7 k − 4  1 3 k ) (which is divisible
by 3). | A1 | Shows using the inductive hypothesis that f ( k + 1 ) is divisible by 3 so must
be considering an expression for f ( k + 1 ) and not just an expression for
f ( k + 1 ) − f ( k ) .
So, if true for n = k this implies true for n = k + 1. True
when n = 0 so therefore true for all integers n 0. | A1
Conclusion, dependent on B1M1M1A1 and no errors seen in their proof.
Do not award this mark if n = 0 not used as base case. However, this mark
can be awarded if n = 1 used as base case and n = 0 considered separately.
Must mention, ‘n = k or P(k)’, ‘n = k + 1 or P(k + 1)’, ‘n = 0 or base case
or P(0)’ (provided P(n) is defined) and ‘n 0or all n oe’ condone ‘all
integers’ but not ‘positive integers’ or similar incorrect statement.
Assume that, when n = k , 1 1  7 n − 1 3 n − 1 is divisible by
3.
That is, f ( k ) = 1 1  7 k − 1 3 k − 1 is divisible by 3.
M1*
Forms correct inductive hypothesis - condone using n e.g.
f ( n ) = 1 1  7 n − 1 3 n − 1 for this mark. Do not need to use f(k) – can just
f ( k + 1 ) − f ( k ) = ( 1 1  7 k + 1 − 1 3 k + 1 − 1 ) − ( 1 1  7 k − 1 3 k − 1 )
= 6  1 1  7 k − 1 2  1 3 k
Considers correct expression for f ( k + 1 ) − f ( k ) and simplifies to an
expression of the form  a  7 k  b  1 3 k - must not be using n for k.
f ( k + 1 ) = f ( k ) + 3 ( 2 2  7 k − 4  1 3 k ) (which is divisible
by 3).
Shows using the inductive hypothesis that f ( k + 1 ) is divisible by 3 so must
be considering an expression for f ( k + 1 ) and not just an expression for
So, if true for n = k this implies true for n = k + 1. True
when n = 0 so therefore true for all integers n 0.
A1