| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2020 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | First-Order Linear Recurrence Relations |
| Difficulty | Standard +0.8 This is a Further Maths question on non-homogeneous first-order recurrence relations requiring the particular integral method (trying Vₙ = an + b) and applying initial conditions. While systematic, it demands more sophisticated technique than standard A-level sequences and involves algebraic manipulation across multiple steps, placing it moderately above average difficulty. |
| Spec | 8.01f First-order recurrence: solve using auxiliary equation and complementary function |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | Complementary Solution is V = A × 2n |
| Answer | Marks |
|---|---|
| n | B1 |
| Answer | Marks |
|---|---|
| [6] | 1.2 |
| Answer | Marks |
|---|---|
| 1.1 | Allow V = an for method mark |
| Answer | Marks | Guidance |
|---|---|---|
| 14 | 2 | 4 |
| 5 | (b) | V = 8 ⇒ A = 5 so V = 5 × 2n– n – 1 |
| Answer | Marks |
|---|---|
| 20 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| 1.1 | soi (or BC) |
Question 5:
5 | (a) | Complementary Solution is V = A × 2n
n
For Particular Solution, try V = an + b
n
Then V = 2V + n ⇒ an + (a + b) = 2an + 2b + n
n + 1 n
Comparing coefficients: a = 2a + 1 and a + b = 2b
⇒ a = b = –1
General Solution is thus V = A × 2n– n – 1
n | B1
M1
A1
M1
A1
B1
[6] | 1.2
1.1a
1.1
1.1
1.1
1.1 | Allow V = an for method mark
n
Substitution and comparing of coefficients
FT GS = CS + PS provided CS has one arbitrary constant and
PS has none (and is a polynomial)
×
14 | 2 | 4 | 6 | 8 | 10 | 12
5 | (b) | V = 8 ⇒ A = 5 so V = 5 × 2n– n – 1
1 n
So V = 5242859
20 | M1
A1
[2] | 3.1a
1.1 | soi (or BC)
accept exact value only.5
\begin{enumerate}[label=(\alph*)]
\item Determine the general solution of the first-order recurrence relation $V _ { n + 1 } = 2 V _ { n } + n$.
\item Given that $V _ { 1 } = 8$, find the exact value of $V _ { 20 }$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2020 Q5 [8]}}