Edexcel F1 2018 June — Question 8 5 marks

Exam BoardEdexcel
ModuleF1 (Further Pure Mathematics 1)
Year2018
SessionJune
Marks5
PaperDownload PDF ↗
TopicProof by induction
TypeProve matrix power formula
DifficultyStandard +0.8 This is a Further Maths proof by induction involving matrix powers with a non-trivial result. The algebraic manipulation required in the inductive step—particularly simplifying the (2,1) entry to show (a^(k+1) - b^(k+1))/(a-b)—demands careful factorization and is more sophisticated than standard induction proofs. However, the structure follows a clear template and the matrix multiplication itself is straightforward due to the triangular form.
Spec4.01a Mathematical induction: construct proofs

  1. Prove by induction that, for \(n \in \mathbb { Z } ^ { + }\)
$$\left( \begin{array} { l l } a & 0 \\ 1 & b \end{array} \right) ^ { n } = \left( \begin{array} { c c } a ^ { n } & 0 \\ \frac { a ^ { n } - b ^ { n } } { a - b } & b ^ { n } \end{array} \right)$$ where \(a\) and \(b\) are constants and \(a \neq b\).

\begin{enumerate}
  \item Prove by induction that, for $n \in \mathbb { Z } ^ { + }$
\end{enumerate}

$$\left( \begin{array} { l l } 
a & 0 \\
1 & b
\end{array} \right) ^ { n } = \left( \begin{array} { c c } 
a ^ { n } & 0 \\
\frac { a ^ { n } - b ^ { n } } { a - b } & b ^ { n }
\end{array} \right)$$

where $a$ and $b$ are constants and $a \neq b$.\\

\hfill \mbox{\textit{Edexcel F1 2018 Q8 [5]}}