| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Topic | Proof by induction |
| Type | Prove summation formula |
| Difficulty | Challenging +1.2 Part (i) is trivial algebraic manipulation to find a=2, b=1. Part (ii) is a standard proof by induction for a sum of powers formula, though the algebraic manipulation in the inductive step is moderately involved due to the cubic terms. This is a typical Further Maths induction question—more algebraically demanding than A-level Core but still follows a well-rehearsed template without requiring novel insight. |
| Spec | 4.01a Mathematical induction: construct proofs4.06a Summation formulae: sum of r, r^2, r^3 |
\begin{enumerate}[label=(\roman*)]
\item Write down the values of the constants $a$ and $b$ for which $m^3 = \frac{1}{6}m^3(am^2 + 2) - \frac{1}{12}m^2(bm)$. [1]
\item Prove by induction that $\sum_{r=1}^{n} r^5 = \frac{1}{6}n^3(n+1)^3 - \frac{1}{12}n^2(n+1)^2$ for all positive integers $n$. [7]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2018 Q8 [8]}}