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1-x-2023-1-x-x-2-2022-a-0-a-1-x-a-2-x-2-a-6067-x-6067-also-if-the-equations-a-n-1-mod-5-a-n-2-mod-5-n-0-1-2-6067-has-u-and-v-solutions-respectivley-then-prove-that-




Question Number 140606 by mathsuji last updated on 10/May/21
(1+x)^(2023) (1−x+x^2 )^(2022) =a_0 +a_1 x+a_2 x^2 +...+a_(6067) x^(6067)   also if the equations  a_n ≡ 1(mod 5) , a_n ≡ 2(mod 5) , n=0;1;2;...;6067  has u and v solutions respectivley,  then prove that...  (√(1+2(√(1+3(√(1+4(√(1+...)))))))) = (√(1+u∙v))
$$\left(\mathrm{1}+{x}\right)^{\mathrm{2023}} \left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)^{\mathrm{2022}} ={a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +…+{a}_{\mathrm{6067}} {x}^{\mathrm{6067}} \\ $$$${also}\:{if}\:{the}\:{equations} \\ $$$${a}_{{n}} \equiv\:\mathrm{1}\left({mod}\:\mathrm{5}\right)\:,\:{a}_{{n}} \equiv\:\mathrm{2}\left({mod}\:\mathrm{5}\right)\:,\:{n}=\mathrm{0};\mathrm{1};\mathrm{2};…;\mathrm{6067} \\ $$$${has}\:{u}\:{and}\:{v}\:{solutions}\:{respectivley}, \\ $$$${then}\:{prove}\:{that}… \\ $$$$\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+…}}}}\:=\:\sqrt{\mathrm{1}+{u}\centerdot{v}} \\ $$
Commented by mathsuji last updated on 10/May/21
Sir mr.W pliz...
$${Sir}\:{mr}.{W}\:{pliz}… \\ $$
Commented by mr W last updated on 10/May/21
(√(1+2(√(1+3(√(1+4(√(1+...)))))))) = 3  a_n =C_k ^(2022)  with n=3k, k=0,1,..,2022  a_n =C_k ^(2022)  with n=3k+1  we have 4 times a_n  with a_n ≡1 mod 5,  i.e. 2 times C_0 ^(2022)  and 2 times C_2 ^(2022) .  we have 2 times a_n  with a_n ≡2 mod 5,  i.e. 2 times C_1 ^(2022) .  that means u=4, v=2  (√(1+uv))=(√(1+4×2))=3
$$\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+…}}}}\:=\:\mathrm{3} \\ $$$${a}_{{n}} ={C}_{{k}} ^{\mathrm{2022}} \:{with}\:{n}=\mathrm{3}{k},\:{k}=\mathrm{0},\mathrm{1},..,\mathrm{2022} \\ $$$${a}_{{n}} ={C}_{{k}} ^{\mathrm{2022}} \:{with}\:{n}=\mathrm{3}{k}+\mathrm{1} \\ $$$${we}\:{have}\:\mathrm{4}\:{times}\:{a}_{{n}} \:{with}\:{a}_{{n}} \equiv\mathrm{1}\:{mod}\:\mathrm{5}, \\ $$$${i}.{e}.\:\mathrm{2}\:{times}\:{C}_{\mathrm{0}} ^{\mathrm{2022}} \:{and}\:\mathrm{2}\:{times}\:{C}_{\mathrm{2}} ^{\mathrm{2022}} . \\ $$$${we}\:{have}\:\mathrm{2}\:{times}\:{a}_{{n}} \:{with}\:{a}_{{n}} \equiv\mathrm{2}\:{mod}\:\mathrm{5}, \\ $$$${i}.{e}.\:\mathrm{2}\:{times}\:{C}_{\mathrm{1}} ^{\mathrm{2022}} . \\ $$$${that}\:{means}\:{u}=\mathrm{4},\:{v}=\mathrm{2} \\ $$$$\sqrt{\mathrm{1}+{uv}}=\sqrt{\mathrm{1}+\mathrm{4}×\mathrm{2}}=\mathrm{3} \\ $$
Commented by mathsuji last updated on 10/May/21
perfect Sir thanks
$${perfect}\:{Sir}\:{thanks} \\ $$

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