OCR Further Additional Pure AS 2019 June — Question 4 8 marks

Exam BoardOCR
ModuleFurther Additional Pure AS (Further Additional Pure AS)
Year2019
SessionJune
Marks8
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Mark schemeDownload PDF ↗
TopicSequences and Series
TypeFirst-Order Linear Recurrence Relations
DifficultyChallenging +1.2 This is a first-order linear recurrence relation with a polynomial forcing term (n²). While it requires knowledge of the particular integral method for non-homogeneous recurrences (trying u_n = an² + bn + c), this is a standard technique taught in Further Maths. The solution involves straightforward algebraic manipulation to find coefficients, making it moderately harder than average but still a textbook-style question.
Spec8.01a Recurrence relations: general sequences, closed form and recurrence8.01f First-order recurrence: solve using auxiliary equation and complementary function
4 The sequence \(\left\{ \mathrm { u } _ { \mathrm { n } } \right\}\) is defined by \(u _ { 1 } = 1\) and \(\mathrm { u } _ { \mathrm { n } + 1 } = 2 \mathrm { u } _ { \mathrm { n } } + \mathrm { n } ^ { 2 }\) for \(\mathrm { n } \geqslant 1\).
Determine \(u _ { n }\) as a function of \(n\).
Question 4:
AnswerMarks
4u – 2u = n2
n + 1 n
Complementary Soln. is u = A2n
n
For Particular Soln., try u = an2 + bn + c
n
Substg. into given r.r. for both u and u
n + 1 n
an2 + 2an + a + bn + b + c – 2( an2 + bn + c) = n2
Comparing coeffts.
a = –1, b = –2, c = –3 so that PS is u = – (n2 + 2n + 3)
n
General Soln. is u = A2n – (n2 + 2n + 3)
n
Use of initial term to evaluate A
u = 1 = 2A – (1 + 2 + 3)  A = 7
1 2
and Soln. is u = 72n – 1 – (n2 + 2n + 3) oe
AnswerMarks
nB1
M1
M1
M1
A1
B1
M1
AnswerMarks
A11.1
2.1
1.1
1.1
2.2a
1.1
1.1a
AnswerMarks
1.1FT
cao If all correct, the final A mark may be
awarded at the previous line
Alternative method
u = A2n + an2 + bn + c
AnswerMarks Guidance
nB1 M1 CS, PS
{u } = {1, 3, 10, 29, 74, …}
AnswerMarks Guidance
nM1 Using correct first four (or five?) terms in system
of equations
1 2Aabc
3 4A4a2bc
Setting up system of equations:
10 8A9a3bc
AnswerMarks Guidance
29 16A16a4bcM1 M1 M1 for at least two; M2 for all four (or five?)
Solving system of equationsM1 BC
A = 7 and a = –1, b = –2, c = –3
AnswerMarks
2A1 A1
[8]
Question 4:
4 | u – 2u = n2
n + 1 n
Complementary Soln. is u = A2n
n
For Particular Soln., try u = an2 + bn + c
n
Substg. into given r.r. for both u and u
n + 1 n
an2 + 2an + a + bn + b + c – 2( an2 + bn + c) = n2
Comparing coeffts.
a = –1, b = –2, c = –3 so that PS is u = – (n2 + 2n + 3)
n
General Soln. is u = A2n – (n2 + 2n + 3)
n
Use of initial term to evaluate A
u = 1 = 2A – (1 + 2 + 3)  A = 7
1 2
and Soln. is u = 72n – 1 – (n2 + 2n + 3) oe
n | B1
M1
M1
M1
A1
B1
M1
A1 | 1.1
2.1
1.1
1.1
2.2a
1.1
1.1a
1.1 | FT
cao If all correct, the final A mark may be
awarded at the previous line
Alternative method
u = A2n + an2 + bn + c
n | B1 M1 | CS, PS
{u } = {1, 3, 10, 29, 74, …}
n | M1 | Using correct first four (or five?) terms in system
of equations
1 2Aabc
3 4A4a2bc
Setting up system of equations:
10 8A9a3bc
29 16A16a4bc | M1 M1 | M1 for at least two; M2 for all four (or five?)
Solving system of equations | M1 | BC
A = 7 and a = –1, b = –2, c = –3
2 | A1 A1
[8]
4 The sequence $\left\{ \mathrm { u } _ { \mathrm { n } } \right\}$ is defined by $u _ { 1 } = 1$ and $\mathrm { u } _ { \mathrm { n } + 1 } = 2 \mathrm { u } _ { \mathrm { n } } + \mathrm { n } ^ { 2 }$ for $\mathrm { n } \geqslant 1$.\\
Determine $u _ { n }$ as a function of $n$.

\hfill \mbox{\textit{OCR Further Additional Pure AS 2019 Q4 [8]}}
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