3 Prove by mathematical induction that, for all non-negative integers \(n\),
$$11 ^ { 2 n } + 25 ^ { n } + 22$$
is divisible by 24 .
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Question 3:
Answer Marks
Guidance
Working/Answer Mark
Guidance
\(H_k\): \(f(k) = 11^{2k} + 25^k + 22 = 24\lambda\) B1
\(f(0) = 1 + 1 + 22 = 24 = 1 \times 24 \Rightarrow H_0\) is true B1
\(f(k+1) - f(k) = 11^{2k+2} + 25^{k+1} + 22 - (11^{2k} + 25^k + 22)\) M1
\(= 11^{2k}(121-1) + 25^k(25-1)\) A1
\(= 11^{2k} \times 24 \times 5 + 25^k \times 24 = 24\mu\) A1
Alternatively: \(f(k+1) = 11^{2k+2} + 25^{k+1} + 22\)(M1)
\(= (120+1)11^{2k} + (24+1)25^k + 22 = 120 \cdot 11^{2k} + 24 \cdot 25^k + 24\lambda = 24\mu\) (A1)(A1)
OE
\(\Rightarrow f(k+1) = 24\mu + 24\lambda = 24(\mu + \lambda) \Rightarrow H_{k+1}\) is true A1
Hence by PMI \(H_n\) is true for all non-negative integers. Must see non-negative integers. CSO: Final mark requires all previous marks.
Total: [6]
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## Question 3:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $H_k$: $f(k) = 11^{2k} + 25^k + 22 = 24\lambda$ | B1 | |
| $f(0) = 1 + 1 + 22 = 24 = 1 \times 24 \Rightarrow H_0$ is true | B1 | |
| $f(k+1) - f(k) = 11^{2k+2} + 25^{k+1} + 22 - (11^{2k} + 25^k + 22)$ | M1 | |
| $= 11^{2k}(121-1) + 25^k(25-1)$ | A1 | |
| $= 11^{2k} \times 24 \times 5 + 25^k \times 24 = 24\mu$ | A1 | |
| **Alternatively:** $f(k+1) = 11^{2k+2} + 25^{k+1} + 22$ | (M1) | |
| $= (120+1)11^{2k} + (24+1)25^k + 22 = 120 \cdot 11^{2k} + 24 \cdot 25^k + 24\lambda = 24\mu$ | (A1)(A1) | OE |
| $\Rightarrow f(k+1) = 24\mu + 24\lambda = 24(\mu + \lambda) \Rightarrow H_{k+1}$ is true | A1 | |
| Hence by PMI $H_n$ is true for all non-negative integers. Must see non-negative integers. CSO: Final mark requires all previous marks. | | |
**Total: [6]**
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3 Prove by mathematical induction that, for all non-negative integers $n$,
$$11 ^ { 2 n } + 25 ^ { n } + 22$$
is divisible by 24 .
\hfill \mbox{\textit{CAIE FP1 2014 Q3 [6]}}