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Question Number 193485 by Socracious last updated on 15/Jun/23

$$$$$$\mathbf{Show}\mathbf{that}{2}^{\mathbf{n}}-{(-1)}^{\mathbf{n}}\mathbf{is}\mathbf{divisible}\mathbf{by}$$$$3\mathbf{for}\mathbf{all}\mathbf{positive}\mathbf{integers}\mathbf{n}.$$

Answered by MM42 last updated on 15/Jun/23

$$k\in \mathbb{N}$$$$ifn=2k;\Rightarrow 2^n-{(-1)}^n=4^k-1\stackrel{3}{\equiv }1-1=0$$$$ifn=2k+1;\Rightarrow 2^n-{(-1)}^n=2\times 4^k+1\stackrel{3}{\equiv }2+1=3\stackrel{3}{\equiv }0$$$$$$

Answered by witcher3 last updated on 15/Jun/23

$${\mathrm{x}}^{\mathrm{n}}-{\mathrm{y}}^{\mathrm{n}}=(\mathrm{x}-\mathrm{y}).\left(\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}-1}}{\mathrm{x}}^{\mathrm{k}}{\mathrm{y}}^{\mathrm{n}-1-\mathrm{k}}\right)$$$$2^{\mathrm{n}}-{(-1)}^{\mathrm{n}}=3.\left(\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}-1}}{2}^{\mathrm{k}}{(-1)}^{\mathrm{n}-1-\mathrm{k}}\right)$$

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