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Question Number 185137 by emmanuelson123 last updated on 17/Jan/23

Answered by CElcedricjunior last updated on 17/Jan/23

$$\mathit{m}\mathit{o}\mathit{n}\mathit{t}\mathit{r}\mathit{o}\mathit{n}\mathit{s}\mathit{q}\mathit{u}\mathit{e}\mathrm{\forall }\mathit{n}\in \mathit{I}\mathit{N}$$$$\frac{\sqrt{3}}{4}[{(7+4\sqrt{3})}^{\mathit{n}}-{(7-4\sqrt{3})}^{\mathit{n}}]\equiv 0\left[6\right]$$$$\mathit{p}\mathit{o}\mathit{s}\mathit{o}\mathit{n}\mathit{s}\mathit{p}\left(\mathit{n}\right)=\frac{\sqrt{3}}{4}[{(7+4\sqrt{3})}^{\mathit{n}}-{(7-4\sqrt{3})}^{\mathit{n}}]$$$$\ast \mathit{p}\mathit{o}\mathit{u}\mathit{r}\mathit{n}=0\mathit{o}\mathit{n}\mathit{a}$$$$\frac{\sqrt{3}}{4}[{(7+4\sqrt{3})}^0-{(7-4\sqrt{3})}^0]=\frac{\sqrt{3}}{4}(1-1)=0\equiv 0\left[6\right]$$$${\mathit{D}}^{\prime }\mathit{o}\stackrel{`}{\mathit{u}}p\left(0\right)\mathit{v}\mathit{r}\mathit{a}\mathit{i}$$$$\ast \mathit{p}\mathit{o}\mathit{u}\mathit{r}\mathit{n}⩾0\mathit{s}\mathit{u}\mathit{p}\mathit{p}\mathit{o}\mathit{s}\mathit{o}\mathit{n}\mathit{s}\mathit{p}\left(\mathit{n}\right)\mathit{v}\mathit{r}\mathit{a}\mathit{i}\mathit{e}\mathit{e}\mathit{t}\mathit{m}\mathit{o}\mathit{n}\mathit{t}\mathit{r}\mathit{o}\mathit{n}\mathit{s}\mathit{q}\mathit{u}\mathit{e}\mathit{p}(\mathit{n}+1){\mathit{l}}^{\prime }\mathit{e}\mathit{s}\mathit{t}\mathit{a}\mathit{u}\mathit{s}\mathit{s}\mathit{i}$$$$\mathit{a}\mathit{u}\mathit{r}\mathit{a}\mathit{n}\mathit{g}(\mathit{n}+1)\mathit{o}\mathit{n}\mathit{a}$$$$\frac{\sqrt{3}}{4}[{(7+4\sqrt{3})}^{\mathit{n}}-{(7-4\sqrt{3})}^{\mathit{n}}]\equiv 0\left[6\right]$$$$=>\frac{\sqrt{3}{(7+4\sqrt{3})}^n}{4}\equiv \frac{\sqrt{3}}{4}{(7-4\sqrt{3})}^{\mathit{n}}\left[6\right]$$$$=>{(7+4\sqrt{3})}^{\mathit{n}}\equiv {(7-4\sqrt{3})}^{\mathit{n}}\left[6\right]$$$$=>{(7+4\sqrt{3})}^{\mathit{n}+1}\equiv {(7-4\sqrt{3})}^{\mathit{n}-1}\left[6\right]$$$$=>{(7+4\sqrt{3})}^{\mathit{n}+1}-{(7-4\sqrt{3})}^{\mathit{n}-1}\equiv 0\left[6\right]$$$$=>\frac{\sqrt{3}}{4}[{(7+4\sqrt{3})}^{\mathit{n}+1}-{(7-4\sqrt{3})}^{\mathit{n}+1}]\equiv 0\left[6\right]\mathit{c}\mathit{a}\mathit{s}{(7-4\sqrt{3})}^{\mathit{n}-1}⩽{(7-4\sqrt{3})}^{\mathit{n}+1}$$$${\mathit{d}}^{\prime }\mathit{o}\stackrel{`}{\mathit{u}}\mathit{p}(\mathit{n}+1)\mathit{v}\mathit{r}\mathit{a}\mathit{i}\mathit{e}$$$$\mathit{d}\mathit{o}\mathit{n}\mathit{c}\mathrm{\forall }\mathit{n}\in \mathit{I}\mathit{N}$$$$\text{Missing left or extra right}$$$$$$$$$$

Commented by aba last updated on 17/Jan/23

$$\mathrm{proof}\mathrm{by}\mathrm{recurence}?$$

Answered by JDamian last updated on 17/Jan/23

$$S_n=\sqrt{3}[{(7+4\sqrt{3})}^n-{(7-4\sqrt{3})}^n]=$$$$=\sqrt{3}\mathrm{\Sigma }[C_{n,k}\cdot (7^{n-k}\cdot {\left(4\sqrt{3}\right)}^k]1⩽k⩽nk=2p+1$$$$S_n=\mathrm{\Sigma }[C_{n,k}\cdot (7^{n-k}\cdot \sqrt{3}\cdot {\left(4\sqrt{3}\right)}^k]$$$$$$$$\sqrt{3}\cdot {\left(4\sqrt{3}\right)}^k=\sqrt{3}\cdot {\left(4\sqrt{3}\right)}^{2p+1}=$$$$=\sqrt{3}\cdot {\left[{\left(4\sqrt{3}\right)}^2\right]}^p\cdot 4\sqrt{3}=$$$$={48}^p\cdot 12$$$$$$$$S_n=\mathrm{\Sigma }(C_{n,k}\cdot 7^{n-k}\cdot {48}^p\cdot 12)$$$$({48}^p\cdot 12)\mathrm{mod}6=0\Rightarrow S_n\mathrm{mod}6=0$$

Commented by aba last updated on 17/Jan/23

$${\mathrm{S}}_{\mathrm{n}}=\frac{\sqrt{3}}{4}....$$

Commented by aba last updated on 17/Jan/23

$${\mathrm{a}}^{\mathrm{n}}-{\mathrm{b}}^{\mathrm{n}}=(\mathrm{a}-\mathrm{b})\left(\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}-1}}{\mathrm{a}}^{\mathrm{k}}{\mathrm{b}}^{\mathrm{n}-1-\mathrm{k}}\right)$$$${\mathrm{S}}_{\mathrm{n}}=\frac{\sqrt{3}}{4}[8\sqrt{3}.\left(\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}-1}}{(7+4\sqrt{3})}^{\mathrm{k}}{(7-4\sqrt{3})}^{\mathrm{n}-1-\mathrm{k}}\right)]$$$$=6\left(\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}-1}}{(\sqrt{3}+2)}^{2\mathrm{k}}{(\sqrt{3}-2)}^{2(\mathrm{n}-1-\mathrm{k})}\right)$$$$\equiv 0\mathrm{mod}\left(6\right)$$$$$$

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