Moderate -0.3 This is a standard proof by induction for divisibility with straightforward algebraic manipulation. While it requires proper induction structure and factoring out 7 from the inductive step, the algebra is routine (factoring 2^{k+2} + 5×9^k from 2^{k+2} + 5×9^{k+1}) and the question follows a well-practiced template. Slightly easier than average due to its predictable structure, though the exponential manipulation requires care.
Basis case. Must explicitly state divisibility (\(49 \div 7 = 7\) is OK). Do not condone \(k=0\) unless later stated \(1>0\)
Assume true for \(n=k\) i.e. that \(2^{k+1}+5\times9^k\) is divisible by 7 oe
M1 (AO 2.1)
Statement of inductive hypothesis. Allow "\(= 7p\)" without further qualification
Considering \(2^{k+1+1}+5\times9^{k+1}\) and rewriting the first term as \(2\times2^{k+1}\) or the second term as \(9\times5\times9^k\)
M1 (AO 1.1)
Might not all be done before next step. Do not allow if e.g. \(45^k\). Needs to have a \(2^{k+1}\) or \(5\times9^k\) so that the \(n=k\) case can be used.
\(2(7p-5\times9^k)+9\times5\times9^k\) or \(2\times2^{k+1}+(7p-2^{k+1})\)
M1 (AO 1.1)
Uses inductive hypothesis properly. Do not allow if e.g. \(45^k\). Do not allow M1 if both replacements made unless recovered later
\(7(2p+5\times9^k)\) or \(7(9p-2^{k+1})\) (which is divisible by 7)
A1 (AO 2.2a)
Simplification with sufficient working to establish divisibility for \(k+1\)
So true for \(n=k \Rightarrow\) true for \(n=k+1\). But true for \(n=1\). So true for all positive integers \(n \geq 1\)
E1 (AO 2.4)
Clear conclusion for induction process. A formal proof by induction is required for full marks.
[6 marks]
# Question 7:
| Answer | Mark | Guidance |
|--------|------|----------|
| $k=1$, 49 is divisible by 7 | M1 (AO 2.1) | Basis case. Must explicitly state divisibility ($49 \div 7 = 7$ is OK). Do not condone $k=0$ unless later stated $1>0$ |
| Assume true for $n=k$ i.e. that $2^{k+1}+5\times9^k$ is divisible by 7 oe | M1 (AO 2.1) | Statement of inductive hypothesis. Allow "$= 7p$" without further qualification |
| Considering $2^{k+1+1}+5\times9^{k+1}$ and rewriting the first term as $2\times2^{k+1}$ or the second term as $9\times5\times9^k$ | M1 (AO 1.1) | Might not all be done before next step. Do not allow if e.g. $45^k$. Needs to have a $2^{k+1}$ or $5\times9^k$ so that the $n=k$ case can be used. |
| $2(7p-5\times9^k)+9\times5\times9^k$ or $2\times2^{k+1}+(7p-2^{k+1})$ | M1 (AO 1.1) | Uses inductive hypothesis properly. Do not allow if e.g. $45^k$. Do not allow M1 if both replacements made unless recovered later |
| $7(2p+5\times9^k)$ or $7(9p-2^{k+1})$ (which is divisible by 7) | A1 (AO 2.2a) | Simplification with sufficient working to establish divisibility for $k+1$ |
| So true for $n=k \Rightarrow$ true for $n=k+1$. But true for $n=1$. So true for all positive integers $n \geq 1$ | E1 (AO 2.4) | Clear conclusion for induction process. A formal proof by induction is required for full marks. |
**[6 marks]**
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7 Prove by induction that $2 ^ { n + 1 } + 5 \times 9 ^ { n }$ is divisible by 7 for all integers $n \geqslant 1$.
\hfill \mbox{\textit{OCR Further Pure Core AS 2018 Q7 [6]}}