OCR MEI Further Extra Pure Specimen — Question 3 12 marks

Exam BoardOCR MEI
ModuleFurther Extra Pure (Further Extra Pure)
SessionSpecimen
Marks12
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Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeRecurrence relation solving for closed form
DifficultyChallenging +1.2 This is a Further Maths recurrence relation question requiring the auxiliary equation method (repeated root λ=4), algebraic manipulation to derive the recursive formula for v_n, limit analysis, and applying the general solution u_n = (A+Bn)4^n to find limits. While it involves multiple parts and some algebraic insight, the techniques are standard for Further Maths students and the question provides significant scaffolding through its structured parts.
Spec4.10d Second order homogeneous: auxiliary equation method8.01d Sequence limits: limit of nth term as n tends to infinity, steady-states8.01f First-order recurrence: solve using auxiliary equation and complementary function
  1. Find the general solution of $$u_n = 8u_{n-1} - 16u_{n-2}, \quad n \geq 2. \quad (*)$$ [4]
A new sequence \(v_n\) is defined by \(v_n = \frac{u_n}{u_{n-1}}\) for \(n \geq 1\).
  1. (A) Use (*) to show that \(v_n = 8 - \frac{16}{v_{n-1}}\) for \(n \geq 2\). [2] (B) Deduce that if \(v_n\) tends to a limit then it must be 4. [2]
  2. Use your general solution in part (i) to show that \(\lim_{n \to \infty} v_n = 4\). [3]
  3. Deduce the value of \(\lim_{n \to \infty} \left(\frac{u_n}{u_{n-2}}\right)\). [1]
Question 3:
AnswerMarks Guidance
3(i) u 8u 16u 0
n n1 n2
28160
Auxiliary equation:
(4)2 0repeated root 4
u (ABn)4 n
AnswerMarks
nM1
A1
M1
A1
AnswerMarks
[4]1.1a
1.1
1.1
AnswerMarks Guidance
1.1Or BC
3(ii) (A)
n n1 n2
u u
n 816 n2
u u
n1 n1
u u 1
n v and n2 
u n u v
n1 n1 n1
16
v 8 AG
n v
AnswerMarks
n1M1
E1
AnswerMarks
[2]1.1
3.1a
AnswerMarks Guidance
3(ii) (B)
If v tends to a limit  then 8 .
n 
28160
which has only one root, 4.
So if v tends to a limit then it must be 4.
AnswerMarks
nM1
E1
AnswerMarks
[2]2.2a
1.1Uniqueness must be
shown
AnswerMarks Guidance
3(iii) u (ABn)4 n
v  n 
n u (AB(n1))4 n1
n1
(AB)4
 n
(AB B)
n n
4B
 as n
B
AnswerMarks
= 4 AGM1
A1
A1
AnswerMarks
[3]3.1a
1.1
AnswerMarks Guidance
2.2aMust be convincing
3(iv) 42 = 16
[1]2.2a 
Question 3:
3 | (i) | u 8u 16u 0
n n1 n2
28160
Auxiliary equation:
(4)2 0repeated root 4
u (ABn)4 n
n | M1
A1
M1
A1
[4] | 1.1a
1.1
1.1
1.1 | Or BC
3 | (ii) | (A) | u 8u 16u , n2
n n1 n2
u u
n 816 n2
u u
n1 n1
u u 1
n v and n2 
u n u v
n1 n1 n1
16
v 8 AG
n v
n1 | M1
E1
[2] | 1.1
3.1a
3 | (ii) | (B) | 16
If v tends to a limit  then 8 .
n 
28160
which has only one root, 4.
So if v tends to a limit then it must be 4.
n | M1
E1
[2] | 2.2a
1.1 | Uniqueness must be
shown
3 | (iii) | u (ABn)4 n
v  n 
n u (AB(n1))4 n1
n1
(AB)4
 n
(AB B)
n n
4B
 as n
B
= 4 AG | M1
A1
A1
[3] | 3.1a
1.1
2.2a | Must be convincing
3 | (iv) | 42 = 16 | B1
[1] | 2.2a
\begin{enumerate}[label=(\roman*)]
\item Find the general solution of
$$u_n = 8u_{n-1} - 16u_{n-2}, \quad n \geq 2. \quad (*)$$ [4]
\end{enumerate}

A new sequence $v_n$ is defined by $v_n = \frac{u_n}{u_{n-1}}$ for $n \geq 1$.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item (A) Use (*) to show that $v_n = 8 - \frac{16}{v_{n-1}}$ for $n \geq 2$. [2]

(B) Deduce that if $v_n$ tends to a limit then it must be 4. [2]

\item Use your general solution in part (i) to show that $\lim_{n \to \infty} v_n = 4$. [3]

\item Deduce the value of $\lim_{n \to \infty} \left(\frac{u_n}{u_{n-2}}\right)$. [1]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Extra Pure  Q3 [12]}}
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