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Question Number 101473 by 175 last updated on 02/Jul/20

Answered by 1549442205 last updated on 03/Jul/20

$$$$$$\mathrm{We}\mathrm{prove}\mathrm{by}\mathrm{induction}\mathrm{method}\mathrm{that}$$$$\mathrm{\exists }\mathrm{m}\in \mathbf{N}\mathrm{such}\mathrm{that}$$$${\mathrm{A}}_{\mathrm{n}}={(2+\sqrt{2})}^{\mathrm{n}}=({\mathrm{m}}_{\mathrm{n}}+\sqrt{{\mathrm{m}}_{\mathrm{n}}^2-2^{\mathrm{n}}})\mathrm{and}{\mathrm{q}}_{\mathrm{n}}=\sqrt{\frac{{\mathrm{m}}_{\mathrm{n}}^2-2^{\mathrm{n}}}{2}}\in \mathbb{N}$$$${\mathrm{A}}_{\mathrm{n}}={\mathrm{m}}_{\mathrm{n}}+{\mathrm{q}}_{\mathrm{n}}\sqrt{2}$$$$+\mathrm{For}\mathrm{n}=1\mathrm{we}\mathrm{have}2+\sqrt{2}=2+\sqrt{2^2-2}$$$$\mathrm{and}\sqrt{\frac{2^2-2}{2}}=1\in \mathbf{N}\Rightarrow \mathrm{State}\mathrm{true}$$$$+\mathrm{Suppose}\mathrm{State}\mathrm{was}\mathrm{true}\mathrm{for}\mathrm{\forall }\mathrm{n}⩽\mathrm{k}$$$$\mathrm{means}\mathrm{that}\mathrm{we}\mathrm{had}{\mathrm{A}}_{\mathrm{k}}={(2+\sqrt{2})}^{\mathrm{k}}=({\mathrm{m}}_{\mathrm{k}}+\sqrt{{\mathrm{m}}_{\mathrm{k}}^2-2^{\mathrm{k}}})$$$$\mathrm{and}{\mathrm{q}}_{\mathrm{k}}=\sqrt{\frac{{\mathrm{m}}_{\mathrm{k}}^2-2^{\mathrm{k}}}{2}}\in \mathbf{N},{\mathrm{A}}_{\mathrm{k}}={\mathrm{m}}_{\mathrm{k}}+{\mathrm{q}}_{\mathrm{k}}\sqrt{2}$$$$+\mathrm{Then}{\mathrm{A}}_{\mathrm{k}+1}={(2+\sqrt{2})}^{\mathrm{k}+1}={(2+\sqrt{2})}^{\mathrm{k}}(2+\sqrt{2})$$$$={\mathrm{A}}_{\mathrm{k}}(2+\sqrt{2})=({\mathrm{m}}_{\mathrm{k}}+\sqrt{{\mathrm{m}}_{\mathrm{k}}^2-2^{\mathrm{k}}})(2+\sqrt{2})$$$$={2\mathrm{m}}_{\mathrm{k}}+\sqrt{{2\mathrm{m}}_{\mathrm{k}}^2-2^{\mathrm{k}+1}}+\sqrt{2}{\mathrm{m}}_{\mathrm{k}}+2\sqrt{{\mathrm{m}}_{\mathrm{k}}^2-2^{\mathrm{k}}}$$$$=({2\mathrm{m}}_{\mathrm{k}}+2\sqrt{\frac{{\mathrm{m}}_{\mathrm{k}}^2-2^{\mathrm{k}}}{2}})+\sqrt{2}({\mathrm{m}}_{\mathrm{k}}+\sqrt{{2\mathrm{m}}_{\mathrm{k}}^2-2^{\mathrm{k}+1}})$$$$={2\mathrm{m}}_{\mathrm{k}}+{2\mathrm{q}}_{\mathrm{k}}+({\mathrm{m}}_{\mathrm{k}}+{2\mathrm{q}}_{\mathrm{k}})\sqrt{2}=2({\mathrm{m}}_{\mathrm{k}}+{\mathrm{q}}_{\mathrm{k}})+\sqrt{2{({\mathrm{m}}_{\mathrm{k}}+{2\mathrm{q}}_{\mathrm{k}})}^2}$$$$=2({\mathrm{m}}_{\mathrm{k}}+{\mathrm{q}}_{\mathrm{k}})+\sqrt{{2\mathrm{m}}_{\mathrm{k}}^2+{8\mathrm{m}}_{\mathrm{k}}{\mathrm{q}}_{\mathrm{k}}+{8\mathrm{q}}_{\mathrm{k}}^2}$$$$=2({\mathrm{m}}_{\mathrm{k}}+{\mathrm{q}}_{\mathrm{k}})+\sqrt{{2\mathrm{m}}_{\mathrm{k}}^2+{8\mathrm{m}}_{\mathrm{k}}{\mathrm{q}}_{\mathrm{k}}+{4\mathrm{q}}_{\mathrm{k}}^2+4.\frac{{\mathrm{m}}_{\mathrm{k}}^2-2^{\mathrm{k}}}{2}}$$$$=2({\mathrm{m}}_{\mathrm{k}}+{\mathrm{q}}_{\mathrm{k}})+\sqrt{({4\mathrm{m}}_{\mathrm{k}}^2+{8\mathrm{m}}_{\mathrm{k}}{\mathrm{q}}_{\mathrm{k}}+{4\mathrm{q}}_{\mathrm{k}}^2)-2^{\mathrm{k}+1}}$$$$=2({\mathrm{m}}_{\mathrm{k}}+{\mathrm{q}}_{\mathrm{k}})+\sqrt{4{({\mathrm{m}}_{\mathrm{k}}+{\mathrm{q}}_{\mathrm{k}})}^2-2^{\mathrm{k}+1}.}$$$$\mathrm{Putting}{\mathrm{m}}_{\mathrm{k}+1}=2({\mathrm{m}}_{\mathrm{k}}+{\mathrm{q}}_{\mathrm{k}}),{\mathrm{q}}_{\mathrm{k}+1}={\mathrm{m}}_{\mathrm{k}}+{2\mathrm{q}}_{\mathrm{k}}$$$$\mathrm{then}{\mathrm{A}}_{\mathrm{k}+1}={\mathrm{m}}_{\mathrm{k}+1}+\sqrt{{\mathrm{m}}_{\mathrm{k}+1}^2-2^{\mathrm{k}+1}}\mathrm{and}$$$${\mathrm{q}}_{\mathrm{k}+1=}\sqrt{\frac{{\mathrm{m}}_{\mathrm{k}+1}^2-2^{\mathrm{k}+1}}{2}}\in \mathbf{N}\mathrm{which}\mathrm{shows}$$$$\mathrm{that}\mathrm{State}\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{n}=\mathrm{k}+1$$$$\mathrm{By}\mathrm{indution}\mathrm{principle}\mathrm{State}\mathrm{is}\mathrm{true}$$$$\mathrm{\forall }\mathrm{n}\in \mathbf{N}.\mathrm{Thus},\mathrm{there}\mathrm{is}\mathrm{always}\mathrm{exists}\mathrm{a}$$$$\mathrm{natural}\mathrm{number}\mathrm{m}\mathrm{so}\mathrm{that}$$$${(2+\sqrt{2})}^{\mathbf{n}}=(\mathbf{m}+\sqrt{{\mathbf{m}}^2-2^{\mathbf{n}}}\mathrm{\forall }\mathbf{n}\in \mathbf{N}(\mathbf{q}.\mathbf{e}.\mathbf{d})$$

Commented by 175 last updated on 03/Jul/20

thank, please answer the quation 2

Commented by 175 last updated on 03/Jul/20

khd

Commented by 1549442205 last updated on 04/Jul/20

$$\mathrm{You}\mathrm{are}\mathrm{welcome}\mathrm{sir}.$$

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