| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Session | Specimen |
| Marks | 13 |
| Topic | Sequences and series, recurrence and convergence |
| Type | Recurrence relation solving for closed form |
| Difficulty | Challenging +1.2 This is a structured multi-part question requiring proof by induction, integration by parts, and connecting sequences to integrals. While it involves several techniques and the connection between parts requires insight, each individual step is fairly standard for Further Maths: the induction follows naturally from the recurrence relation, the integration by parts is routine, and the final connection is guided by the question structure. The difficulty is elevated above average due to the multi-step nature and need to synthesize different areas, but it's well within reach for a competent Further Maths student following the scaffolding provided. |
| Spec | 4.01a Mathematical induction: construct proofs4.08f Integrate using partial fractions8.01b Induction: prove results for sequences and series8.06a Reduction formulae: establish, use, and evaluate recursively |
**(i)** For $n=0$, $u_0 = 1 = 0! \times \frac{1}{0!}$ so statement is true for $n=0$: B1
(Allow for those who show it is true for $n=1$)
Assuming $u_k = k!\sum_{r=0}^{k}\frac{1}{r!}$, for induction hypothesis: B1
$u_{k+1} = (k+1)u_k + 1$, for use of recurrence relation to obtain $u_{k+1}$: M1
$= (k+1)!\sum_{r=0}^{k}\frac{1}{r!} + \frac{(k+1)!}{(k+1)!}$
$= (k+1)!\sum_{r=0}^{k+1}\frac{1}{r!}$: A1
which is the formula with $n = k+1$
Hence, **if** it is true for $n=k$ then it is also true for $n=k+1$
Since true for $n=0$ ($n=1$), it is true for $n=2, n=3, \ldots$, i.e. all integers $n \geq 0$ (1)
For valid explanation of the inductive logic: B1 **[5]**
**(ii)(a)** $I_n = \int_0^1 x^n e^{-x}\,dx = \left[x^n(-e^{-x})\right]_0^1 - \int_0^1 nx^{n-1}(-e^{-x})\,dx$, for use of parts: M1, A1
$= nI_{n-1} - \frac{1}{e}$, correctly shown: A1 **[3]**
**(ii)(b)** $I_0 = \int_0^1 e^{-x}\,dx = \left[-e^{-x}\right]_0^1 = 1 - \frac{1}{e}$: B1
$I_1 = I_0 - \frac{1}{e} = 1! - \frac{2}{e}$ from reduction formula: B1 **[2]**
**(ii)(c)** Assuming $I_k = k! - \frac{u_k}{e}$, for an attempt at an inductive proof: M1
(allow base case implicitly)
$I_{k+1} = (k+1)\left(k! - \frac{u_k}{e}\right) - \frac{1}{e} = (k+1)! - \frac{(k+1)u_k + 1}{e}$
For dealing with the sequence numerator: M1(dep)
$= (k+1)! - \frac{u_{k+1}}{e}$ (no need for formal conclusion): A1 **[3]**12
\begin{enumerate}[label=(\roman*)]
\item The sequence $\left\{ u _ { n } \right\}$ is defined for all integers $n \geq 0$ by
$$u _ { 0 } = 1 \quad \text { and } \quad u _ { n } = n u _ { n - 1 } + 1 , \quad n \geq 1 .$$
Prove by induction that $u _ { n } = n ! \sum _ { r = 0 } ^ { n } \frac { 1 } { r ! }$.
\item (a) Given that $I _ { n } = \int _ { 0 } ^ { 1 } x ^ { n } \mathrm { e } ^ { - x } \mathrm {~d} x$ for $n \geq 0$, show that, for $n \geq 1$,
$$I _ { n } = n I _ { n - 1 } - \frac { 1 } { \mathrm { e } }$$
(b) Evaluate $I _ { 0 }$ exactly and deduce the value of $I _ { 1 }$.\\
(c) Show that $I _ { n } = n ! - \frac { u _ { n } } { \mathrm { e } }$ for all integers $n \geq 1$.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 Q12 [13]}}