Challenging +1.2 This is a standard reduction formula question with a straightforward induction proof. The first part requires integration by parts (a routine technique), and the induction step follows naturally from the reduction formula. While it involves multiple components, each step is methodical and typical for Further Pure 1, making it moderately above average difficulty but not requiring novel insight.
7 Let
$$I _ { n } = \int _ { 0 } ^ { 1 } t ^ { n } \mathrm { e } ^ { - t } \mathrm {~d} t$$
where \(n \geqslant 0\). Show that, for all \(n \geqslant 1\),
$$I _ { n } = n I _ { n - 1 } - \mathrm { e } ^ { - 1 }$$
Hence prove by induction that, for all positive integers \(n\),
$$I _ { n } < n ! .$$
(Do not allow \(I_0\) unless \(I_1\) is deduced from it)
Completes induction argument
A1
This requires at least: 'The assumption is true and it is true for all positive integers' for the final mark, which is only awarded if all previous marks have been gained in the proof part.
$I_n = \left[-t^ne^{-t}\right]_0^n + n\int_0^t t^{n-1}e^{-t}dt$ | M1A1 |
$I_n = nI_{n-1} - e^{-1}$ (AG) | A1 |
$H_k: I_k < k!$ for some $k$ | B1 |
$H_k \Rightarrow I_{k+1} < (k+1)k! - e^{-1} < (k+1)!$ | M1A1 |
$I_1 = 1 - 2e^{-1}$ | B1 |
(Do not allow $I_0$ unless $I_1$ is deduced from it) |
Completes induction argument | A1 |
**This requires at least:** 'The assumption is true and it is true for all positive integers' for the final mark, which is only awarded if all previous marks have been gained in the proof part. |
7 Let
$$I _ { n } = \int _ { 0 } ^ { 1 } t ^ { n } \mathrm { e } ^ { - t } \mathrm {~d} t$$
where $n \geqslant 0$. Show that, for all $n \geqslant 1$,
$$I _ { n } = n I _ { n - 1 } - \mathrm { e } ^ { - 1 }$$
Hence prove by induction that, for all positive integers $n$,
$$I _ { n } < n ! .$$
\hfill \mbox{\textit{CAIE FP1 2009 Q7 [8]}}