OCR MEI Further Extra Pure 2022 June — Question 3 9 marks

Exam BoardOCR MEI
ModuleFurther Extra Pure (Further Extra Pure)
Year2022
SessionJune
Marks9
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Mark schemeDownload PDF ↗
TopicSequences and Series
TypeFirst-Order Linear Recurrence Relations
DifficultyChallenging +1.8 This is a Further Maths question requiring solution of a non-homogeneous first-order linear recurrence with quadratic forcing term, followed by limit analysis. While the techniques are standard for Further Maths students (complementary function + particular integral, then asymptotic behavior), the quadratic forcing term and multi-part limit analysis requiring careful algebraic manipulation place it well above average difficulty, though not at the extreme end for Further Pure modules.
Spec8.01d Sequence limits: limit of nth term as n tends to infinity, steady-states8.01f First-order recurrence: solve using auxiliary equation and complementary function
3 A sequence is defined by the recurrence relation \(5 t _ { n + 1 } - 4 t _ { n } = 3 n ^ { 2 } + 28 n + 6\), for \(n \geqslant 0\), with \(t _ { 0 } = 7\).
  1. Find an expression for \(t _ { n }\) in terms of \(n\). Another sequence is defined by \(\mathrm { v } _ { \mathrm { n } } = \frac { \mathrm { t } _ { \mathrm { n } } } { \mathrm { n } ^ { \mathrm { m } } }\), for \(n \geqslant 1\), where \(m\) is a constant.
  2. In each of the following cases determine \(\lim _ { n \rightarrow \infty } \mathrm {~V} _ { n }\), if it exists, or show that the sequence is divergent.
    1. \(m = 3\)
    2. \(m = 2\)
    3. \(m = 1\)
Question 3:
AnswerMarks Guidance
3(a) CF: 5t – 4t = 0 ⇒t = α(4/5)n
n+1 n n
PS: Try t = an2 + bn + c
n
5(a(n +1)2 + b(n +1) + c) – 4(an2 + bn + c) =
3n2 + 28n + 6 or
5(an2 + bn + c) – 4(a(n –1) 2 + b(n –1) + c) =
3(n –1) 2 + 28(n –1) + 6 oe
an2 + (10a + b)n + 5a + 5b + c ≡ 3n2 + 28n +
6
a = 3, 10a + b = 28, 5a + 5b + c = 6
a = 3, b = –2, c = 1 so
Gen Sol: t = α(4/5)n + 3n2 – 2n + 1
n
t = 7 =>α = 6 ⇒t = 6(4/5)n + 3n2 – 2n + 1
AnswerMarks
0 nB1
M1
M1
M1
A1FT
B1FT
AnswerMarks
[6]1.1
1.1
1.1
1.1
1.1
AnswerMarks
1.1Correct general form for particular
solution
Substituting their form correctly
into recurrence relation.
Expanding, collecting terms and
comparing coefficients
Their CF + correct PS
Substituting t = 7 into their CF +
0
PS solution with a single arbitrary
constant to derive a solution with a
correct first term
AnswerMarks Guidance
3(b) (i)
If m = 3, v = + − +
n n3 n n2 n3
so lim v exists and is equal to 0
n
AnswerMarks Guidance
n→∞B1FT
[1]2.2a Some justification (eg expressing
v in the form shown) must be
n
AnswerMarks
given.For parts (i), (ii) and (iii) marks
can be gained for answers
correctly derived from solutions of
the form
t = αrn + an2 + bn + c
n
provided that r <1.
AnswerMarks Guidance
3(b) (ii)
If m = 2, v =3+ − +
n n2 n n2
so lim v exists and is equal to “3”
n
AnswerMarks Guidance
n→∞B1FT
[1]2.2a Their a (see guidance above).
Some justification (eg expressing
v in the form shown) must be
n
given.
AnswerMarks Guidance
3(b) (iii)
If m = 1, v =3n−2+ +
n
n n
so lim v does not exist
n
AnswerMarks Guidance
n→∞B1FT
[1]2.2a Some justification (eg expressing
v in the form shown) must be
n
AnswerMarks
given.Allow “infinity” or “infinite” or ∞
or “diverges” or “no limit” or
“increases without bound” etc
Question 3:
3 | (a) | CF: 5t – 4t = 0 ⇒t = α(4/5)n
n+1 n n
PS: Try t = an2 + bn + c
n
5(a(n +1)2 + b(n +1) + c) – 4(an2 + bn + c) =
3n2 + 28n + 6 or
5(an2 + bn + c) – 4(a(n –1) 2 + b(n –1) + c) =
3(n –1) 2 + 28(n –1) + 6 oe
an2 + (10a + b)n + 5a + 5b + c ≡ 3n2 + 28n +
6
a = 3, 10a + b = 28, 5a + 5b + c = 6
a = 3, b = –2, c = 1 so
Gen Sol: t = α(4/5)n + 3n2 – 2n + 1
n
t = 7 =>α = 6 ⇒t = 6(4/5)n + 3n2 – 2n + 1
0 n | B1
M1
M1
M1
A1FT
B1FT
[6] | 1.1
1.1
1.1
1.1
1.1
1.1 | Correct general form for particular
solution
Substituting their form correctly
into recurrence relation.
Expanding, collecting terms and
comparing coefficients
Their CF + correct PS
Substituting t = 7 into their CF +
0
PS solution with a single arbitrary
constant to derive a solution with a
correct first term
3 | (b) | (i) | 6×0.8n 3 2 1
If m = 3, v = + − +
n n3 n n2 n3
so lim v exists and is equal to 0
n
n→∞ | B1FT
[1] | 2.2a | Some justification (eg expressing
v in the form shown) must be
n
given. | For parts (i), (ii) and (iii) marks
can be gained for answers
correctly derived from solutions of
the form
t = αrn + an2 + bn + c
n
provided that r <1.
3 | (b) | (ii) | 6×0.8n 2 1
If m = 2, v =3+ − +
n n2 n n2
so lim v exists and is equal to “3”
n
n→∞ | B1FT
[1] | 2.2a | Their a (see guidance above).
Some justification (eg expressing
v in the form shown) must be
n
given.
3 | (b) | (iii) | 6×0.8n 1
If m = 1, v =3n−2+ +
n
n n
so lim v does not exist
n
n→∞ | B1FT
[1] | 2.2a | Some justification (eg expressing
v in the form shown) must be
n
given. | Allow “infinity” or “infinite” or ∞
or “diverges” or “no limit” or
“increases without bound” etc
3 A sequence is defined by the recurrence relation $5 t _ { n + 1 } - 4 t _ { n } = 3 n ^ { 2 } + 28 n + 6$, for $n \geqslant 0$, with $t _ { 0 } = 7$.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $t _ { n }$ in terms of $n$.

Another sequence is defined by $\mathrm { v } _ { \mathrm { n } } = \frac { \mathrm { t } _ { \mathrm { n } } } { \mathrm { n } ^ { \mathrm { m } } }$, for $n \geqslant 1$, where $m$ is a constant.
\item In each of the following cases determine $\lim _ { n \rightarrow \infty } \mathrm {~V} _ { n }$, if it exists, or show that the sequence is divergent.
\begin{enumerate}[label=(\roman*)]
\item $m = 3$
\item $m = 2$
\item $m = 1$
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Extra Pure 2022 Q3 [9]}}
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