| Exam Board | Edexcel |
|---|---|
| Module | FD2 AS (Further Decision 2 AS) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dynamic Programming |
| Type | Recurrence relation asymptotic behaviour |
| Difficulty | Challenging +1.8 This Further Maths question requires substituting a quadratic ansatz into a recurrence relation, equating coefficients of powers of n, and solving a system of equations. While the technique is systematic once recognized, it demands algebraic manipulation across multiple steps and understanding of asymptotic behavior—significantly harder than standard A-level recurrence relations but still a structured problem type in FP2/FD2. |
| Spec | 1.04e Sequences: nth term and recurrence relations4.06b Method of differences: telescoping series |
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| \cline { 3 - 5 } \multicolumn{2}{c|}{} | Team B | |||
| \cline { 3 - 5 } \multicolumn{2}{c|}{} | Paul | Qaasim | Rashid | |
| \multirow{3}{*}{Team A} | Mischa | 4 | - 6 | 2 |
| \cline { 2 - 5 } | Noel | 0 | - 2 | 6 |
| \cline { 2 - 5 } | Olive | - 6 | 2 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Auxiliary equation \(2m-1=0\), complementary function is \(A\left(\frac{1}{2}\right)^n\) | B1 | cao |
| Particular solution try \(u_n = \lambda n^2 + \beta n + \alpha\) and substitute into recurrence relation | M1 | Correct form for particular solution and substituted into recurrence relation |
| \(2\lambda n^2+2\beta n+2\alpha=(\lambda-k)n^2+(-2\lambda+\beta)n+(\lambda-\beta+\alpha)\) \(\Rightarrow 2\lambda=\lambda-k\), \(2\beta=-2\lambda+\beta\), \(2\alpha=\lambda-\beta+\alpha\) | M1 | Compares coefficients and setting up all three equations in \(\lambda, \beta, \alpha\) |
| \(u_n = A\left(\frac{1}{2}\right)^n - kn^2 + 2kn - 3k\) | A1 | Correct general solution (or with consistent value of \(k\)) |
| \(u_0=A-3k,\ u_2=\frac{1}{4}A-3k \Rightarrow 4\left(\frac{1}{4}A-3k\right)-(A-3k)=27k^2\) | M1 | Use initial condition to obtain a quadratic equation in \(k\) |
| \(27k^2+9k=0 \Rightarrow k=-\frac{1}{3}\ (k\neq 0)\) | A1ft | Correct solution for \(k\) following through their general solution |
| As \(n\) becomes large \(A\left(\frac{1}{2}\right)^n \to 0\) | B1 | Correct explanation that exponential term tends to zero as \(n\) becomes large |
| \(u_n \to \frac{1}{3}n^2 - \frac{2}{3}n+1\ \left(a=\frac{1}{3},\ b=-\frac{2}{3},\ c=1\right)\) | A1 | cao |
## Question 4:
| Answer | Mark | Guidance |
|--------|------|----------|
| Auxiliary equation $2m-1=0$, complementary function is $A\left(\frac{1}{2}\right)^n$ | B1 | cao |
| Particular solution try $u_n = \lambda n^2 + \beta n + \alpha$ and substitute into recurrence relation | M1 | Correct form for particular solution and substituted into recurrence relation |
| $2\lambda n^2+2\beta n+2\alpha=(\lambda-k)n^2+(-2\lambda+\beta)n+(\lambda-\beta+\alpha)$ $\Rightarrow 2\lambda=\lambda-k$, $2\beta=-2\lambda+\beta$, $2\alpha=\lambda-\beta+\alpha$ | M1 | Compares coefficients and setting up all three equations in $\lambda, \beta, \alpha$ |
| $u_n = A\left(\frac{1}{2}\right)^n - kn^2 + 2kn - 3k$ | A1 | Correct general solution (or with consistent value of $k$) |
| $u_0=A-3k,\ u_2=\frac{1}{4}A-3k \Rightarrow 4\left(\frac{1}{4}A-3k\right)-(A-3k)=27k^2$ | M1 | Use initial condition to obtain a quadratic equation in $k$ |
| $27k^2+9k=0 \Rightarrow k=-\frac{1}{3}\ (k\neq 0)$ | A1ft | Correct solution for $k$ following through their general solution |
| As $n$ becomes large $A\left(\frac{1}{2}\right)^n \to 0$ | B1 | Correct explanation that exponential term tends to zero as $n$ becomes large |
| $u_n \to \frac{1}{3}n^2 - \frac{2}{3}n+1\ \left(a=\frac{1}{3},\ b=-\frac{2}{3},\ c=1\right)$ | A1 | cao |4. A sequence $\left\{ u _ { n } \right\}$, where $n \geqslant 1$, satisfies the recurrence relation
$$2 u _ { n } = u _ { n - 1 } - k n ^ { 2 } \text { where } 4 u _ { 2 } - u _ { 0 } = 27 k ^ { 2 }$$
and $k$ is a non-zero constant.\\
Show that, as $n$ becomes large, $u _ { n }$ can be approximated by a quadratic function of the form $a n ^ { 2 } + b n + c$ where $a , b$ and $c$ are constants to be determined.
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\section*{Thursday 14 May 2020}
You may not need to use all of these tables.
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3.
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4.
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\hfill \mbox{\textit{Edexcel FD2 AS 2020 Q4 [8]}}