AQA Further Paper 1 2023 June — Question 5 6 marks

Exam BoardAQA
ModuleFurther Paper 1 (Further Paper 1)
Year2023
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeMethod of differences with given identity
DifficultyStandard +0.8 This is a Further Maths question requiring algebraic manipulation to verify a difference formula, then applying the method of differences to sum a series. Part (a) is straightforward algebra (2 marks), but part (b) requires understanding telescoping sums and careful bookkeeping with powers of 2. While the method of differences is a standard Further Maths technique, executing it correctly with the exponential terms and simplifying to the given form requires more sophistication than typical A-level questions, placing it moderately above average difficulty.
Spec4.06b Method of differences: telescoping series

The function f is defined by $$f(r) = 2^r(r - 2) \quad (r \in \mathbb{Z})$$
  1. Show that $$f(r + 1) - f(r) = r2^r$$ [2 marks]
  2. Use the method of differences to show that $$\sum_{r=1}^n r2^r = 2^{n+1}(n - 1) + 2$$ [4 marks]

Question 5:

AnswerMarks
5(a)Obtains a correct expression for
f (r + 1) and forms an
expression for the difference
f(r+1)−f(r)
AnswerMarks Guidance
Eg 2r+1( r−1 )−2r( r−2 )1.1a M1
=2r( 2 ( r−1 )−( r−2 ))
=2r( 2r−2−r+2 )
=r2r
Completes a rigorous argument
to obtain the required result
Must include f(r+1)−f(r), r2r
and at least one intermediate
step between
AnswerMarks Guidance
2r+1( r−1 )−2r( r−2 ) and r2r2.1 R1
Subtotal2
QMarking instructions AO

AnswerMarks
5(b)Writes at least two lines of
subtracting terms, using part (a)
AnswerMarks Guidance
Allow f(2)−f(1)etc1.1a M1
∑ r2r =∑ f(r+1)−f(r)
r=1 r=1
= 0 −(−2 )
+ 8 − 0
+32 −8
+...
+2n( n−2 ) − 2n−1( n−3 )
+2n+1( n−1 )− 2n( n−2 )
n
∑ r2r =2n+1( n−1 )−2 (−1 )
r=1
=2n+1( n−1 )+2
Writes at least two consecutive
AnswerMarks Guidance
lines showing cancellation (PI)2.5 M1
Correctly reduces the
AnswerMarks Guidance
expression to two terms1.1b A1
Completes a reasoned
argument using the method of
differences to reach the required
result.
This mark is only available if at
least the first two lines and the
last two lines, to reach the
AnswerMarks Guidance
required result, are seen.2.1 R1
Subtotal4
Question total6
QMarking instructions AO
Question 5:
--- 5(a) ---
5(a) | Obtains a correct expression for
f (r + 1) and forms an
expression for the difference
f(r+1)−f(r)
Eg 2r+1( r−1 )−2r( r−2 ) | 1.1a | M1 | f(r+1)−f(r)=2r+1( r−1 )−2r( r−2 )
=2r( 2 ( r−1 )−( r−2 ))
=2r( 2r−2−r+2 )
=r2r
Completes a rigorous argument
to obtain the required result
Must include f(r+1)−f(r), r2r
and at least one intermediate
step between
2r+1( r−1 )−2r( r−2 ) and r2r | 2.1 | R1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 5(b) ---
5(b) | Writes at least two lines of
subtracting terms, using part (a)
Allow f(2)−f(1)etc | 1.1a | M1 | n n
∑ r2r =∑ f(r+1)−f(r)
r=1 r=1
= 0 −(−2 )
+ 8 − 0
+32 −8
+...
+2n( n−2 ) − 2n−1( n−3 )
+2n+1( n−1 )− 2n( n−2 )
n
∑ r2r =2n+1( n−1 )−2 (−1 )
r=1
=2n+1( n−1 )+2
Writes at least two consecutive
lines showing cancellation (PI) | 2.5 | M1
Correctly reduces the
expression to two terms | 1.1b | A1
Completes a reasoned
argument using the method of
differences to reach the required
result.
This mark is only available if at
least the first two lines and the
last two lines, to reach the
required result, are seen. | 2.1 | R1
Subtotal | 4
Question total | 6
Q | Marking instructions | AO | Marks | Typical solution
The function f is defined by
$$f(r) = 2^r(r - 2) \quad (r \in \mathbb{Z})$$

\begin{enumerate}[label=(\alph*)]
\item Show that
$$f(r + 1) - f(r) = r2^r$$
[2 marks]

\item Use the method of differences to show that
$$\sum_{r=1}^n r2^r = 2^{n+1}(n - 1) + 2$$
[4 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 1 2023 Q5 [6]}}