Moderate -0.3 This is a straightforward proof by induction with a simple divisibility statement. The base case is trivial (4×1+66=70=14×5), and the inductive step requires only basic algebraic manipulation to factor out 14. While it's a Further Maths topic, it's a standard textbook exercise requiring no novel insight—slightly easier than average due to its mechanical nature.
Basis case. Must explicitly see 70 and state divisibility. If \(n=1\) used and not corrected, allow mark but withhold final mark
Assume true for \(n=k\): \(4\times8^k + 66\) is divisible by 14
M1
Statement of inductive hypothesis. Allow "\(= 14p\)" without further qualification
Considering \(4\times8^{k+1} + 66\), rewriting as \(4\times8\times8^k + 66\) or \(32\times8^k + 66\)
M1
Uses law of indices correctly to obtain expression in terms of \(8^k\) with no other exponential term
\(= 8(14p - 66) + 66\) from inductive hypothesis
M1
Uses inductive hypothesis properly. Do not allow if e.g. \(32^k\) used (law of indices must be correctly used)
\(= 112p - 462\) or \(14\times8p - 14\times33 = 14(8p-33)\), divisible by 14
A1
Must either show \(14\times\) explicitly in each term or 14 is a factor
So true for \(n=k \Rightarrow\) true for \(n=k+1\). True for \(n=0\). So true for all integers \(n \geq 0\)
A1
Clear conclusion. Need to see if true for \(n=k\), then true for \(n=k+1\), i.e. \(k\) and \(k+1\)th case linked to \(n\)
# Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $n=0$: $4\times1 + 66 = 70 = 5\times14$, divisible by 14 | **B1** | Basis case. Must explicitly see 70 and state divisibility. If $n=1$ used and not corrected, allow mark but withhold final mark |
| Assume true for $n=k$: $4\times8^k + 66$ is divisible by 14 | **M1** | Statement of inductive hypothesis. Allow "$= 14p$" without further qualification |
| Considering $4\times8^{k+1} + 66$, rewriting as $4\times8\times8^k + 66$ or $32\times8^k + 66$ | **M1** | Uses law of indices correctly to obtain expression in terms of $8^k$ with no other exponential term |
| $= 8(14p - 66) + 66$ from inductive hypothesis | **M1** | Uses inductive hypothesis properly. Do not allow if e.g. $32^k$ used (law of indices must be correctly used) |
| $= 112p - 462$ or $14\times8p - 14\times33 = 14(8p-33)$, divisible by 14 | **A1** | Must either show $14\times$ explicitly in each term or 14 is a factor |
| So true for $n=k \Rightarrow$ true for $n=k+1$. True for $n=0$. So true for all integers $n \geq 0$ | **A1** | Clear conclusion. Need to see if true for $n=k$, then true for $n=k+1$, i.e. $k$ and $k+1$th case linked to $n$ |
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6 Prove by induction that $4 \times 8 ^ { n } + 66$ is divisible by 14 for all integers $n \geqslant 0$.
\hfill \mbox{\textit{OCR Further Pure Core AS 2023 Q6 [6]}}