5. Prove, by induction, that \(3 ^ { 2 n } + 7\) is divisible by 8 for all positive integers \(n\).
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Question 5:
Answer Marks
Guidance
Answer/Working Marks
Guidance
\(f(1) = 3^2 + 7 = 16 = 8 \times 2\), true for \(n=1\) B1
\(f(1) = 3^2 + 7 = 16\) seen
Assume true for \(n=k\): \(f(k) = 3^{2k}+7 = 8p\) \(f(k+1) - f(k) = 3^{2(k+1)}+7-(3^{2k}+7)\) M1
Substituting into \(f(k+1)-f(k)\) or showing \(f(k+1) = 9\times 3^{2k}+7\)
\(= 9\times 3^{2k} + 7 - 3^{2k} - 7\) dM1
Using \(f(k+1)-f(k)\) or equivalent
\(= 8\times 3^{2k}\) A1
\(f(k+1) = f(k) + 8\times 3^{2k}\) or equivalent
\(f(k+1) = 8(3^{2k}+p) = 8q\) dM1
Showing divisible by 8: \(f(k)\) divisible by 8 and \(8\times 3^{2k}\) divisible by 8
True for \(n=k+1\); true for \(n=1\), if true for \(n=k\) then true for \(n=k+1\); so \(3^{2n}+7\) divisible by 8 for all \(n\) by induction A1cso
All 4 underlined elements seen anywhere in solution
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# Question 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(1) = 3^2 + 7 = 16 = 8 \times 2$, true for $n=1$ | B1 | $f(1) = 3^2 + 7 = 16$ seen |
| Assume true for $n=k$: $f(k) = 3^{2k}+7 = 8p$ | | |
| $f(k+1) - f(k) = 3^{2(k+1)}+7-(3^{2k}+7)$ | M1 | Substituting into $f(k+1)-f(k)$ or showing $f(k+1) = 9\times 3^{2k}+7$ |
| $= 9\times 3^{2k} + 7 - 3^{2k} - 7$ | dM1 | Using $f(k+1)-f(k)$ or equivalent |
| $= 8\times 3^{2k}$ | A1 | $f(k+1) = f(k) + 8\times 3^{2k}$ or equivalent |
| $f(k+1) = 8(3^{2k}+p) = 8q$ | dM1 | Showing divisible by 8: $f(k)$ divisible by 8 and $8\times 3^{2k}$ divisible by 8 |
| True for $n=k+1$; true for $n=1$, if true for $n=k$ then true for $n=k+1$; so $3^{2n}+7$ divisible by 8 for all $n$ by induction | A1cso | All 4 underlined elements seen anywhere in solution |
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5. Prove, by induction, that $3 ^ { 2 n } + 7$ is divisible by 8 for all positive integers $n$.\\
\hfill \mbox{\textit{Edexcel FP1 2013 Q5 [6]}}