| Exam Board | Edexcel |
|---|---|
| Module | F1 (Further Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove summation formula |
| Difficulty | Standard +0.3 This is a standard Further Maths induction question with two routine parts: (i) proving a summation formula requiring algebraic manipulation of a cubic expression, and (ii) matrix powers requiring matrix multiplication. Both follow the standard induction template with no novel insights required, making it slightly easier than average even for Further Maths. |
| Spec | 4.01a Mathematical induction: construct proofs4.03c Matrix multiplication: properties (associative, not commutative) |
\begin{enumerate}
\item (i) Prove by induction that, for $n \in \mathbb { Z } ^ { + }$,
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } r ^ { 2 } ( 2 r - 1 ) = \frac { 1 } { 6 } n ( n + 1 ) \left( 3 n ^ { 2 } + n - 1 \right)$$
(ii) Prove by induction that, for $n \in \mathbb { Z } ^ { + }$,
$$\left( \begin{array} { c c }
7 & - 12 \\
3 & - 5
\end{array} \right) ^ { n } = \left( \begin{array} { c c }
6 n + 1 & - 12 n \\
3 n & 1 - 6 n
\end{array} \right)$$
\hfill \mbox{\textit{Edexcel F1 2015 Q9 [12]}}