Pre-U Pre-U 9795 Specimen — Question 13

Exam BoardPre-U
ModulePre-U 9795 (Pre-U Further Mathematics)
SessionSpecimen
TopicReduction Formulae
TypeProve formula by induction
DifficultyChallenging +1.8 This is a challenging Further Maths question requiring integration by parts to derive a reduction formula, then a non-trivial induction proof involving the relationship between I_n and J_n. The induction step requires algebraic manipulation of the reduction formula combined with factorial relationships, which is conceptually demanding but follows established techniques for this topic.
Spec4.01a Mathematical induction: construct proofs

13 Let \(I _ { n } = \int _ { 1 } ^ { \mathrm { e } } ( \ln x ) ^ { n } \mathrm {~d} x\), where \(n\) is a positive integer.
  1. By considering \(\frac { \mathrm { d } } { \mathrm { d } x } \left( x ( \ln x ) ^ { n } \right)\), or otherwise, show that \(I _ { n } = \mathrm { e } - n I _ { n - 1 }\).
  2. Let \(J _ { n } = \frac { I _ { n } } { n ! }\). Prove by induction that $$\sum _ { r = 2 } ^ { n } \frac { ( - 1 ) ^ { r } } { r ! } = \frac { 1 } { \mathrm { e } } \left( 1 + ( - 1 ) ^ { n } J _ { n } \right)$$ for all positive integers \(n \geqslant 2\).

13 Let $I _ { n } = \int _ { 1 } ^ { \mathrm { e } } ( \ln x ) ^ { n } \mathrm {~d} x$, where $n$ is a positive integer.\\
(i) By considering $\frac { \mathrm { d } } { \mathrm { d } x } \left( x ( \ln x ) ^ { n } \right)$, or otherwise, show that $I _ { n } = \mathrm { e } - n I _ { n - 1 }$.\\
(ii) Let $J _ { n } = \frac { I _ { n } } { n ! }$. Prove by induction that

$$\sum _ { r = 2 } ^ { n } \frac { ( - 1 ) ^ { r } } { r ! } = \frac { 1 } { \mathrm { e } } \left( 1 + ( - 1 ) ^ { n } J _ { n } \right)$$

for all positive integers $n \geqslant 2$.

\hfill \mbox{\textit{Pre-U Pre-U 9795  Q13}}