Question Number 140606 by mathsuji last updated on 10/May/21
$$\left(\mathrm{1}+{x}\right)^{\mathrm{2023}} \left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)^{\mathrm{2022}} ={a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +...+{a}_{\mathrm{6067}} {x}^{\mathrm{6067}} \\ $$$${also}\:{if}\:{the}\:{equations} \\ $$$${a}_{{n}} \equiv\:\mathrm{1}\left({mod}\:\mathrm{5}\right)\:,\:{a}_{{n}} \equiv\:\mathrm{2}\left({mod}\:\mathrm{5}\right)\:,\:{n}=\mathrm{0};\mathrm{1};\mathrm{2};...;\mathrm{6067} \\ $$$${has}\:{u}\:{and}\:{v}\:{solutions}\:{respectivley}, \\ $$$${then}\:{prove}\:{that}... \\ $$$$\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+...}}}}\:=\:\sqrt{\mathrm{1}+{u}\centerdot{v}} \\ $$
Commented by mathsuji last updated on 10/May/21
$${Sir}\:{mr}.{W}\:{pliz}... \\ $$
Commented by mr W last updated on 10/May/21
$$\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+...}}}}\:=\:\mathrm{3} \\ $$$${a}_{{n}} ={C}_{{k}} ^{\mathrm{2022}} \:{with}\:{n}=\mathrm{3}{k},\:{k}=\mathrm{0},\mathrm{1},..,\mathrm{2022} \\ $$$${a}_{{n}} ={C}_{{k}} ^{\mathrm{2022}} \:{with}\:{n}=\mathrm{3}{k}+\mathrm{1} \\ $$$${we}\:{have}\:\mathrm{4}\:{times}\:{a}_{{n}} \:{with}\:{a}_{{n}} \equiv\mathrm{1}\:{mod}\:\mathrm{5}, \\ $$$${i}.{e}.\:\mathrm{2}\:{times}\:{C}_{\mathrm{0}} ^{\mathrm{2022}} \:{and}\:\mathrm{2}\:{times}\:{C}_{\mathrm{2}} ^{\mathrm{2022}} . \\ $$$${we}\:{have}\:\mathrm{2}\:{times}\:{a}_{{n}} \:{with}\:{a}_{{n}} \equiv\mathrm{2}\:{mod}\:\mathrm{5}, \\ $$$${i}.{e}.\:\mathrm{2}\:{times}\:{C}_{\mathrm{1}} ^{\mathrm{2022}} . \\ $$$${that}\:{means}\:{u}=\mathrm{4},\:{v}=\mathrm{2} \\ $$$$\sqrt{\mathrm{1}+{uv}}=\sqrt{\mathrm{1}+\mathrm{4}×\mathrm{2}}=\mathrm{3} \\ $$
Commented by mathsuji last updated on 10/May/21
$${perfect}\:{Sir}\:{thanks} \\ $$