Standard +0.3 This is a standard proof by induction with straightforward algebra. The base case is trivial (n=1 gives 27+128=155=19×8+17... wait, let me recalculate: 3^4+2^7=81+128=209=19×11). The inductive step requires factoring out 19 from 3^3·f(k)-2^3·(previous term), which is a routine manipulation once you recognize the pattern. While it's Further Maths content, it's a textbook induction question with no novel insight required, making it slightly easier than average overall.
The function \(f\) is defined by
$$f(n) = 3^{3n+1} + 2^{3n+4} \quad (n \in \mathbb{Z}^+)$$
Prove by induction that \(f(n)\) is divisible by 19 for \(n \geq 1\)
[6 marks]
Question 12:
12 | Shows that f(n)is divisible by 19
for n =1 | 1.1b | B1 | Let n =1; then the formula gives
f(1)=3 4 +2 7 =209=11×19
so the result is true forn =1
Assume the result is true for n= k,so
f(k)=3 3k+1+2 3k+4 =19m ( m∈)
Then
f(k+1)=3 3k+4 +23k+7
( 3k+1) ( 3k+4)
=27 3 +8 2
( 3k+1) ( 3k+1) ( 3k+4)
=19 3 +8 3 +8 2
( 3k+1)
=19 3 +8f(k)
=19 ( 3 3k+1+8m )
and the result also holds for n=k+1
f(n)is divisible by 19 for n=1; if true
for n=k , then it’s also true for
n=k+1 and hence by induction f(n)
is divisible by 19 for n≥1
States the assumption that f(n)
is divisible by 19 for n=k | 2.4 | M1
Expresses f(k+1)in terms of k | 3.1a | M1
Expresses f(k+1)or
f(k+1)−f(k)in the form
a ( 33k+1 ) +b ( 23k+4 ) | 3.1a | M1
Completes reasoned working to
correctly deduce that
f(k+1)is divisible by 19 | 2.2a | R1
Concludes a reasoned
argument by stating that f(n)is
divisible by 19 for n=1; if true
for n=k, then it’s also true for
n=k+1 and hence by induction
f(n)is divisible by 19 for n≥1 | 2.1 | R1
Question total | 6
Q | Marking Instructions | AO | Marks | Typical solution
The function $f$ is defined by
$$f(n) = 3^{3n+1} + 2^{3n+4} \quad (n \in \mathbb{Z}^+)$$
Prove by induction that $f(n)$ is divisible by 19 for $n \geq 1$
[6 marks]
\hfill \mbox{\textit{AQA Further Paper 2 2023 Q12 [6]}}