Standard +0.3 This is a standard two-part induction question from Further Pure 1. Part (i) involves proving a summation formula with straightforward algebraic manipulation, while part (ii) is a divisibility proof requiring modular arithmetic. Both follow routine induction templates with no novel insights required, making this slightly easier than average for Further Maths students but appropriately challenging for the syllabus.
9. (i) Prove by induction that, for \(n \in \mathbb { Z } ^ { + }\)
$$\sum _ { r = 1 } ^ { n } r ( r + 1 ) ( r + 2 ) = \frac { n ( n + 1 ) ( n + 2 ) ( n + 3 ) } { 4 }$$
(ii) Prove by induction that,
$$4 ^ { n } + 6 n + 8 \text { is divisible by } 18$$
for all positive integers \(n\).
\includegraphics[max width=\textwidth, alt={}, center]{df5ab400-5cb1-4b51-8b0a-52dc3587f81a-16_62_44_2476_1889}
9. (i) Prove by induction that, for $n \in \mathbb { Z } ^ { + }$
$$\sum _ { r = 1 } ^ { n } r ( r + 1 ) ( r + 2 ) = \frac { n ( n + 1 ) ( n + 2 ) ( n + 3 ) } { 4 }$$
(ii) Prove by induction that,
$$4 ^ { n } + 6 n + 8 \text { is divisible by } 18$$
for all positive integers $n$.\\
\includegraphics[max width=\textwidth, alt={}, center]{df5ab400-5cb1-4b51-8b0a-52dc3587f81a-16_62_44_2476_1889}
\hfill \mbox{\textit{Edexcel F1 2014 Q9 [11]}}