Standard +0.3 This is a standard proof by induction for divisibility, requiring verification of the base case (n=1) and an inductive step showing 2^(3(k+1)) - 3^(k+1) is divisible by 5 given 2^(3k) - 3^k is. The algebraic manipulation (factoring out 8 from 2^(3k+3) and handling the 3^(k+1) term) is routine for Further Maths students familiar with induction proofs, making it slightly easier than average overall difficulty.
1.1 — Uses laws of indices and the inductive hypothesis properly to eliminate either \(2^{3k}\) or \(3^k\) (not both). Or \(8\times2^{3k}-3\times(2^{3k}-5p)\)
\(= 5\times8p+5\times3^k = 5(8p+3^k) = 5q\) for some integer \(q\), so this is also a multiple of \(5\)
A1
2.2a — AG. Further simplification to establish truth for \(k+1\). \(5(3p+2^{3k})\)
So true for \(n=k \Rightarrow\) true for \(n=k+1\). But true for \(n=1\). So true for all integers \(n\geq1\)
A1
2.4 — Clear conclusion for induction process, following a correct proof by induction. A formal proof by induction is required for full marks.
# Question 7:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Basis Case: when $n=1$: $2^{3n}-3^n = 2^3-3 = 8-3=5$, which is divisible by $5$. | B1 | 2.1 — At least one intermediate step must be shown |
| Assume true for $n=k$ i.e. $2^{3k}-3^k=5p$ for some integer $p$ | M1 | 2.1 — Must have statement in terms of some other variable than $n$ |
| $2^{3(k+1)}-3^{k+1} = 2^3\times2^{3k}-3\times3^k = 8(5p+3^k)-3\times3^k$ | M1 | 1.1 — Uses laws of indices and the inductive hypothesis properly to eliminate either $2^{3k}$ or $3^k$ (not both). Or $8\times2^{3k}-3\times(2^{3k}-5p)$ |
| $= 5\times8p+5\times3^k = 5(8p+3^k) = 5q$ for some integer $q$, so this is also a multiple of $5$ | A1 | 2.2a — AG. Further simplification to establish truth for $k+1$. $5(3p+2^{3k})$ |
| So true for $n=k \Rightarrow$ true for $n=k+1$. But true for $n=1$. So true for all integers $n\geq1$ | A1 | 2.4 — Clear conclusion for induction process, following a **correct** proof by induction. A formal proof by induction is required for full marks. |
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