| Exam Board | OCR MEI |
|---|---|
| Module | Further Extra Pure (Further Extra Pure) |
| Year | 2021 |
| Session | November |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Recurrence relation solving for closed form |
| Difficulty | Challenging +1.3 This is a standard Further Maths recurrence relation question requiring the auxiliary equation method for non-homogeneous relations. Parts (a)-(c) are routine textbook exercises (finding complementary function, particular integral of form An+B, applying initial conditions). Part (d) adds mild complexity by requiring identification that the dominant term determines convergence (p=5) and evaluation of the limit, but this is still a predictable extension. The question is harder than average A-level due to being Further Maths content, but straightforward within that context. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (a) | CF: u – 3u – 10u = 0 and u = αrn |
| Answer | Marks |
|---|---|
| n | M1 |
| Answer | Marks |
|---|---|
| [6] | 1.1a |
| Answer | Marks |
|---|---|
| 1.1 | Deriving the auxiliary equation |
| Answer | Marks |
|---|---|
| must be seen. | Condone missing brackets around |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (b) | Either: n = 0 => 1 + α + β = 6 |
| Answer | Marks |
|---|---|
| n | M1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Substituting n = 0 or n = 1 in their |
| Answer | Marks |
|---|---|
| Do not ISW. | This mark can be awarded if one of |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (c) | From recurrence relation: |
| Answer | Marks | Guidance |
|---|---|---|
| = 1 – 4 + 75 + 8 = 80 | B1 | |
| [1] | 2.5 | Both expressions properly seen (ie |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (d) | 1−2n (5)n (−2)n |
| Answer | Marks |
|---|---|
| q = 3 | M1 |
| Answer | Marks |
|---|---|
| [4] | 3.1a |
| Answer | Marks |
|---|---|
| 2.2a | Writing v in a form which enables |
| Answer | Marks |
|---|---|
| FT. q = α(must be a number). | At most one of c and d is 0. |
Question 4:
4 | (a) | CF: u – 3u – 10u = 0 and u = αrn
n+2 n+1 n n
⇒r2 – 3r – 10 = 0
⇒r = 5 or r = –2
CF is α5n + β(–2)n
Trial function: u = an + b
n
a(n + 2) + b – 3[a(n + 1) + b] – 10(an + b)
= 24n – 10
⇒ (a – 3a – 10a) = 24
and 2a + b – 3a – 3b – 10b = –10
a = –2 and b = 1 so GS is
u = 1 – 2n + α5n + β(–2)n
n | M1
A1FT
B1
M1
M1
A1
[6] | 1.1a
1.1
1.1a
1.1
1.1
1.1 | Deriving the auxiliary equation
(allow one sign error).
FT correct roots of their AE to
form CF (do not ISW).
Correct form.
Substituting their form correctly
into recurrence relation.
Deriving two equations in a and b
using a correct method (eg
comparing coefficients)
Full form of GS, including u =,
n
must be seen. | Condone missing brackets around
–2 unless misused.
Other forms eg an2 + bn + c are
allowable provided a = 0 derived.
cao
4 | (b) | Either: n = 0 => 1 + α + β = 6
or: n = 1 => 1 – 2 + 5α – 2β = 10
α + β = 5 and 5α – 2β = 11
=> 2α + 2β = 10 => 7α = 21
α = 3 and β = 2 so
u = 1 – 2n + 3×5n + 2×(–2)n
n | M1
M1
A1FT
[3] | 1.1
1.1
1.1 | Substituting n = 0 or n = 1 in their
GS to derive an equation in α & β.
Deriving 2 equations from
substituting n = 0 & 1, at least one
correct for their GS, and
attempting to solve .
FT from their GS. Allow non-
embedded values if GS seen in (a).
Do not ISW. | This mark can be awarded if one of
their equations is wrong.
Attempt to solve can be implied by
correct answer or valid algebra but
incorrect answer with no working
M0
4 | (c) | From recurrence relation:
u = 3u + 10u + 24×0 – 10
2 1 0
= 3×10 + 10×6 – 10 = 80
From particular solution:
u = 1 – 2×2 + 3×52 + 2×(–2)2
2
= 1 – 4 + 75 + 8 = 80 | B1
[1] | 2.5 | Both expressions properly seen (ie
it must be clear that candidates are
correctly using two different
methods to find u ).
2
4 | (d) | 1−2n (5)n (−2)n
v = +3 +2
n pn p p
If p <5 then v n →∞ while if p >5 then
v → 0 as n→∞
n
p = 5
q = 3 | M1
B1
A1
A1
[4] | 3.1a
2.1
2.2a
2.2a | Writing v in a form which enables
n
the limit to be deduced.
Convincing argument.
FT for GS of the form:
c – dn + αsn + βtn
(where s > t ).
FT. p = s (must be a number).
FT. q = α(must be a number). | At most one of c and d is 0.
s and t are not equal and both not 0.
Both α and β are not 0.
Either s >1 or t >1 (or both).
A0 If s = –t.
A0 If s = –t.
If M0 then SC2 for p = 5, q = 3.The sequence $u_0, u_1, u_2, \ldots$ satisfies the recurrence relation $u_{n+2} - 3u_{n+1} - 10u_n = 24n - 10$.
\begin{enumerate}[label=(\alph*)]
\item Determine the general solution of the recurrence relation. [6]
\item Hence determine the particular solution of the recurrence relation for which $u_0 = 6$ and $u_1 = 10$. [3]
\item Show, by direct calculation, that your solution in part (b) gives the correct value for $u_2$. [1]
\end{enumerate}
The sequence $v_0, v_1, v_2, \ldots$ is defined by $v_n = \frac{u_n}{p^n}$ for some constant $p$, where $u_n$ denotes the particular solution found in part (b).
You are given that $v_n$ converges to a finite non-zero limit, $q$, as $n \to \infty$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Determine $p$ and $q$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Extra Pure 2021 Q4 [14]}}