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Exercise 4.4 · Q14

Q.Use induction to prove that 10n+3×4n+2+510^n+3\times 4^{n+2}+5, is divisible by 9, for all natural numbers nn.

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Let P(n):10n+3×4n+2+5P(n):10^n+3\times4^{n+2}+5 is divisible by 99.

Step 1. Base case. P(1)P(1): 10+3×43+5=10+192+5=207=9×2310+3\times4^3+5=10+192+5=207=9\times23. True.

Step 2. Inductive hypothesis. Assume P(k):10k+3×4k+2+5=9λP(k):10^k+3\times4^{k+2}+5=9\lambda for some integer λ\lambda, i.e. 10k=9λ−3×4k+2−510^k=9\lambda-3\times4^{k+2}-5. …

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