Moderate -0.3 This is a standard proof by induction for divisibility with a straightforward algebraic manipulation. The inductive step requires factoring 5^(k+1) - 2^(k+1) = 5·5^k - 2·2^k = 5(5^k - 2^k) + 3·2^k, which clearly shows divisibility by 3. While it's a Further Maths topic, this is a routine textbook exercise requiring only standard technique with no novel insight, making it slightly easier than an average A-level question.
Concludes: states \(5^n-2^n\) is divisible by 3 when \(n=1\) (or \(n=0\)), and if true for \(n=k\) then true for \(n=k+1\), hence by induction \(5^n-2^n\) is divisible by 3 for all \(n\in\mathbb{N}\); must define \(m\) to be an integer; condone reference to 'rule'/'statement' in final statement
R1 (2.1)
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## Question 12:
Writes $5^1 - 2^1 = 3$; accept $5^0 - 2^0 = 0$ | B1 (2.1) | —
Assumes $5^k - 2^k$ is divisible by 3 **and** considers $5^{k+1} - 2^{k+1}$; so $5^k = 3m + 2^k$ for some integer $m$ | M1 (2.4) | —
Completes rigorous working: $5^{k+1}-2^{k+1} = 5\times5^k - 2\times2^k = 5(3m+2^k)-2\times2^k = 15m+5\times2^k-2\times2^k = 15m+3\times2^k = 3(5m+2^k) = 3\times\text{integer}$, so divisible by 3 | A1 (2.2a) | —
Concludes: states $5^n-2^n$ is divisible by 3 when $n=1$ (or $n=0$), **and** if true for $n=k$ then true for $n=k+1$, **hence** by induction $5^n-2^n$ is divisible by 3 for all $n\in\mathbb{N}$; must define $m$ to be an integer; condone reference to 'rule'/'statement' in final statement | R1 (2.1) | —
12 Prove by induction that, for all $n \in \mathbb { N }$, the expression
$$5 ^ { n } - 2 ^ { n }$$
is divisible by 3\\[0pt]
[4 marks]\\
LL\\
\hfill \mbox{\textit{AQA Further AS Paper 1 2024 Q12 [4]}}