| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove recurrence relation formula |
| Difficulty | Standard +0.3 This is a straightforward proof by induction with a simple recurrence relation. Part (a) requires basic substitution, and part (b) follows a standard induction template with algebraic manipulation that's simpler than average (just substituting the recurrence relation and factoring out 2^k). The formula is also easy to verify and the inductive step is mechanical, making this slightly easier than a typical A-level question. |
| Spec | 1.04e Sequences: nth term and recurrence relations4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(u_2 = 3\), \(u_3 = 7\) | B1, B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| At \(n=1\), \(u_1 = 2^1 - 1 = 1\), so result true for \(n=1\) | B1 | |
| Assume true for \(n=k\): \(u_k = 2^k - 1\) | — | |
| \(u_{k+1} = 2u_k + 1 = 2(2^k - 1) + 1\) | M1 | Substitutes \(u_k\) into \(u_{k+1}\) (must see this line) |
| Correct expression | A1 | |
| \(u_{k+1} = 2^{k+1} - 2 + 1 = 2^{k+1} - 1\) | A1 | Correct completion to \(u_{k+1} = 2^{k+1}-1\) |
| Must see: true for \(n=1\), assumption true for \(n=k\), said true for \(n=k+1\), therefore true for all \(n\) | A1cso | Fully complete proof with no errors and comment. All previous marks in (b) must have been scored |
## Question 7(a):
| Working | Mark | Guidance |
|---------|------|----------|
| $u_2 = 3$, $u_3 = 7$ | B1, B1 | |
**(2 marks)**
## Question 7(b):
| Working | Mark | Guidance |
|---------|------|----------|
| At $n=1$, $u_1 = 2^1 - 1 = 1$, so result true for $n=1$ | B1 | |
| Assume true for $n=k$: $u_k = 2^k - 1$ | — | |
| $u_{k+1} = 2u_k + 1 = 2(2^k - 1) + 1$ | M1 | Substitutes $u_k$ into $u_{k+1}$ (must see this line) |
| Correct expression | A1 | |
| $u_{k+1} = 2^{k+1} - 2 + 1 = 2^{k+1} - 1$ | A1 | Correct completion to $u_{k+1} = 2^{k+1}-1$ |
| Must see: true for $n=1$, assumption true for $n=k$, said true for $n=k+1$, therefore true for all $n$ | A1cso | Fully complete proof with no errors and comment. All previous marks in (b) must have been scored |
**(5 marks) Total: 7**
---7. A sequence can be described by the recurrence formula
$$u _ { n + 1 } = 2 u _ { n } + 1 , \quad n \geqslant 1 , \quad u _ { 1 } = 1$$
\begin{enumerate}[label=(\alph*)]
\item Find $u _ { 2 }$ and $u _ { 3 }$.
\item Prove by induction that $u _ { n } = 2 ^ { n } - 1$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2012 Q7 [7]}}