Question Number 220700 by universe last updated on 17/May/25
Answered by Frix last updated on 17/May/25
$${a}_{\mathrm{0}} =\sqrt{\mathrm{6}}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}} \\ $$$${a}_{\mathrm{1}} =\sqrt{\mathrm{3}} \\ $$$${a}_{\mathrm{2}} =−\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$${a}_{\mathrm{3}} =−\mathrm{2}+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${a}_{\mathrm{4}} =−\mathrm{2}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${a}_{\mathrm{5}} ={a}_{\mathrm{3}} \\ $$$${a}_{\mathrm{6}} ={a}_{\mathrm{5}} \\ $$$$\Rightarrow \\ $$$$ \\ $$$${n}\geqslant\mathrm{2}:\:\begin{cases}{{a}_{\mathrm{2}{n}−\mathrm{1}} =−\mathrm{2}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\\{{a}_{\mathrm{2}{n}} =−\mathrm{2}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\end{cases} \\ $$