Moderate -0.3 This is a straightforward proof by induction with a given formula to prove. The recurrence relation is simple, and the algebraic manipulation required (expanding and simplifying quadratics) is routine. While it's a Further Maths question, it's a standard textbook induction exercise requiring no novel insight—slightly easier than average for A-level overall.
The sequence \(u_1, u_2, u_3, \ldots\) is defined by
$$u_1 = 1$$
$$u_{n+1} = u_n + 3n$$
Prove by induction that for all integers \(n \geq 1\)
$$u_n = \frac{1}{2}n^2 - \frac{3}{2}n + 1$$
[4 marks]
Question 6:
6 | Shows that
3 3
u = ×12 – ×1+1=1
1
2 2 | 1.1b | B1 | Let n = 1; then the formula gives
3 3
u = × 1 2 – × 1 + 1 = 1
1
2 2
(So the formula is true for n = 1)
Assume the formula is true
for n = k
Then we want to show that
3 3
u = ( k + 1 2 ) – k ( + 1 ) + 1
k + 1
2 2
3
= ( k 2 + 2 k + 1 – k – 1 ) + 1
2
3
= ( k 2 + k ) + 1
2
But
3 3
u = k 2 – k + 1 + 3 k
k + 1
2 2
3
= ( k 2 + k ) + 1
2
as required
So the formula is also true for
n = k + 1
The formula for u n is true for n = 1
If the formula is true for n = k, then
the formula is also true
for n = k + 1
Hence by induction
3 3
u = n 2 – n + 1 is true for all
n
2 2
integers n ≥ 1
Assumes the result is true for
n = k and states
3 3
u = k 2 – k + 1 + 3 k
k + 1
2 2
at any point | 3.1a | M1
Completes correct working to
deduce that
3 3
k 2 – k + 1 + 3 k
2 2
and
3 3
( k + 1 ) 2 – ( k + 1 ) + 1
2 2
are equivalent | 2.2a | B1
Concludes a reasoned
argument by stating:
3 3
u = n 2 – n + 1 true for n = 1
n
2 2
(seen anywhere)
3 3
If u = n2 – n+1
n
2 2
Is true for n = k,
then true for n = k + 1
Hence (by induction)
3 3
u = n 2 – n + 1 is true for all
n
2 2
integers n ≥ 1
Condone “the result/formula”
3 3
instead of u = n 2 – n + 1
n
2 2 | 2.1 | R1
Question total | 4
Q | Marking instructions | AO | Marks | Typical solution
The sequence $u_1, u_2, u_3, \ldots$ is defined by
$$u_1 = 1$$
$$u_{n+1} = u_n + 3n$$
Prove by induction that for all integers $n \geq 1$
$$u_n = \frac{1}{2}n^2 - \frac{3}{2}n + 1$$
[4 marks]
\hfill \mbox{\textit{AQA Further Paper 1 2024 Q6 [4]}}