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Question Number 193485 by Socracious last updated on 15/Jun/23
             Show that 2^n −(−1)^n  is divisible by         3 for all positive integers n.
$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\boldsymbol{\mathrm{Show}}\:\boldsymbol{\mathrm{that}}\:\mathrm{2}^{\boldsymbol{\mathrm{n}}} −\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} \:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{divisible}}\:\boldsymbol{\mathrm{by}} \\ $$$$\:\:\:\:\:\:\:\mathrm{3}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{all}}\:\boldsymbol{\mathrm{positive}}\:\boldsymbol{\mathrm{integers}}\:\boldsymbol{\mathrm{n}}. \\ $$
Answered by MM42 last updated on 15/Jun/23
k∈N  if  n=2k ; ⇒2^n −(−1)^n =4^k −1≡^3 1−1=0  if  n=2k+1 ; ⇒2^n −(−1)^n =2×4^k +1≡^3 2+1=3≡^3 0
$${k}\in\mathbb{N} \\ $$$${if}\:\:{n}=\mathrm{2}{k}\:;\:\Rightarrow\mathrm{2}^{{n}} −\left(−\mathrm{1}\right)^{{n}} =\mathrm{4}^{{k}} −\mathrm{1}\overset{\mathrm{3}} {\equiv}\mathrm{1}−\mathrm{1}=\mathrm{0} \\ $$$${if}\:\:{n}=\mathrm{2}{k}+\mathrm{1}\:;\:\Rightarrow\mathrm{2}^{{n}} −\left(−\mathrm{1}\right)^{{n}} =\mathrm{2}×\mathrm{4}^{{k}} +\mathrm{1}\overset{\mathrm{3}} {\equiv}\mathrm{2}+\mathrm{1}=\mathrm{3}\overset{\mathrm{3}} {\equiv}\mathrm{0} \\ $$$$ \\ $$
Answered by witcher3 last updated on 15/Jun/23
x^n −y^n =(x−y).(Σ_(k=0) ^(n−1) x^k y^(n−1−k) )  2^n −(−1)^n =3.(Σ_(k=0) ^(n−1) 2^k (−1)^(n−1−k) )
$$\mathrm{x}^{\mathrm{n}} −\mathrm{y}^{\mathrm{n}} =\left(\mathrm{x}−\mathrm{y}\right).\left(\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\mathrm{x}^{\mathrm{k}} \mathrm{y}^{\mathrm{n}−\mathrm{1}−\mathrm{k}} \right) \\ $$$$\mathrm{2}^{\mathrm{n}} −\left(−\mathrm{1}\right)^{\mathrm{n}} =\mathrm{3}.\left(\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\mathrm{2}^{\mathrm{k}} \left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}−\mathrm{k}} \right) \\ $$

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