Edexcel FD2 AS 2019 June — Question 2 6 marks

Exam BoardEdexcel
ModuleFD2 AS (Further Decision 2 AS)
Year2019
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDynamic Programming
TypeRecurrence relation solution
DifficultyStandard +0.8 This is a first-order linear non-homogeneous recurrence relation requiring the complementary function plus particular integral method. While the technique is standard for FD2, it demands careful algebraic manipulation to find the particular solution form (trying u_n = k·2^n), then applying the initial condition u_1 = u_2 requires solving for the constant. This is moderately challenging as it tests both procedural fluency and algebraic accuracy across multiple steps, placing it above average difficulty.
Spec1.04e Sequences: nth term and recurrence relations

2. (a) Find the general solution of the recurrence relation $$u _ { n + 1 } = 3 u _ { n } + 2 ^ { n } \quad n \geqslant 1$$ (b) Find the particular solution of this recurrence relation for which \(u _ { 1 } = u _ { 2 }\)

Question 2:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Auxiliary equation \(m - 3 = 0 \Rightarrow\) complementary function is \(A(3)^n\)B1 CAO for complementary function
Consider trial solution \(u_n = k(2^n)\), so \(2k(2^n) = 3k(2^n) + 2^n\)M1 Substituting correct trial solution into recurrence relation; allow substitution of \(u_n = k(2^n)\) into \(u_n = 3u_{n-1} + 2^{n-1}\) but not \(u_n = 3u_{n-1} + 2^n\)
\(k = -1\)A1 CAO
General solution is \(u_n = A(3)^n - 2^n\)A1 CAO for general solution — must include \(u_n = \ldots\)
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(u_1 = u_2 \Rightarrow 3A - 2 = 9A - 4 \Rightarrow A = \ldots\)M1 Using condition \(u_1 = u_2\) to find \(A\ (= \tfrac{1}{3})\); dependent on general solution being of form \(\pm\lambda(3)^n \pm \mu(2)^n\)
\(u_n = 3^{n-1} - 2^n\)A1 CAO; must include \(u_n = \ldots\); if neither general nor particular solution given in terms of \(u_n\), award if correct expression in terms of \(n\) seen
# Question 2:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Auxiliary equation $m - 3 = 0 \Rightarrow$ complementary function is $A(3)^n$ | B1 | CAO for complementary function |
| Consider trial solution $u_n = k(2^n)$, so $2k(2^n) = 3k(2^n) + 2^n$ | M1 | Substituting correct trial solution into recurrence relation; allow substitution of $u_n = k(2^n)$ into $u_n = 3u_{n-1} + 2^{n-1}$ but not $u_n = 3u_{n-1} + 2^n$ |
| $k = -1$ | A1 | CAO |
| General solution is $u_n = A(3)^n - 2^n$ | A1 | CAO for general solution — must include $u_n = \ldots$ |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $u_1 = u_2 \Rightarrow 3A - 2 = 9A - 4 \Rightarrow A = \ldots$ | M1 | Using condition $u_1 = u_2$ to find $A\ (= \tfrac{1}{3})$; dependent on general solution being of form $\pm\lambda(3)^n \pm \mu(2)^n$ |
| $u_n = 3^{n-1} - 2^n$ | A1 | CAO; must include $u_n = \ldots$; if neither general nor particular solution given in terms of $u_n$, award if correct expression in terms of $n$ seen |

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2. (a) Find the general solution of the recurrence relation

$$u _ { n + 1 } = 3 u _ { n } + 2 ^ { n } \quad n \geqslant 1$$

(b) Find the particular solution of this recurrence relation for which $u _ { 1 } = u _ { 2 }$\\

\hfill \mbox{\textit{Edexcel FD2 AS 2019 Q2 [6]}}