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Question-153704

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Question Number 153704 by mathdanisur last updated on 09/Sep/21
Commented by mathdanisur last updated on 09/Sep/21
p_k = (1/(1+x+x^2 +...+x^(2021) )) = ((1 - x)/(1 - x^(k+1) )) = ((1 - x)/(1 - x^(k+1) )) = (1 - x)(1 + (x^(k+1) /(1 - x^(k+1) ))) = ((2020)/(2021)) (1 + (1/(2021^(k+1) - 1))) β‡’Ξ£_(k=1) ^(2021) p_k = ((2020)/(2021)) (2021 + 𝛂) β–² 𝛂 = (1/(2021^2 - 1)) + (1/(2021^3 - 1)) + ... + (1/(2021^(2022) - 1)) < ((2021)/(2021^2 - 1)) = 𝛃 β†’ say β‡’Ξ£_(k=1) ^(2021) p_k = 2020 + ((2020)/(2021)) 𝛃 = 2020 β–²
$$\mathrm{p}_{\boldsymbol{\mathrm{k}}} \:=\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} +…+\mathrm{x}^{\mathrm{2021}} }\:=\:\frac{\mathrm{1}\:-\:\mathrm{x}}{\mathrm{1}\:-\:\mathrm{x}^{\boldsymbol{\mathrm{k}}+\mathrm{1}} } \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{1}\:-\:\mathrm{x}}{\mathrm{1}\:-\:\mathrm{x}^{\boldsymbol{\mathrm{k}}+\mathrm{1}} } \\ $$$$\:\:\:\:\:=\:\left(\mathrm{1}\:-\:\mathrm{x}\right)\left(\mathrm{1}\:+\:\frac{\mathrm{x}^{\boldsymbol{\mathrm{k}}+\mathrm{1}} }{\mathrm{1}\:-\:\mathrm{x}^{\boldsymbol{\mathrm{k}}+\mathrm{1}} }\right) \\ $$$$\:\:\:\:\:=\:\frac{\mathrm{2020}}{\mathrm{2021}}\:\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2021}^{\boldsymbol{\mathrm{k}}+\mathrm{1}} \:-\:\mathrm{1}}\right) \\ $$$$\Rightarrow\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\mathrm{2021}} {\sum}}\mathrm{p}_{\boldsymbol{\mathrm{k}}} \:=\:\frac{\mathrm{2020}}{\mathrm{2021}}\:\left(\mathrm{2021}\:+\:\boldsymbol{\alpha}\right) \\ $$$$\blacktriangle\:\boldsymbol{\alpha}\:=\:\frac{\mathrm{1}}{\mathrm{2021}^{\mathrm{2}} \:-\:\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{2021}^{\mathrm{3}} \:-\:\mathrm{1}}\:+\:…\:+\:\frac{\mathrm{1}}{\mathrm{2021}^{\mathrm{2022}} \:-\:\mathrm{1}}\:<\:\frac{\mathrm{2021}}{\mathrm{2021}^{\mathrm{2}} \:-\:\mathrm{1}}\:=\:\boldsymbol{\beta}\:\rightarrow\:\mathrm{say} \\ $$$$\Rightarrow\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\mathrm{2021}} {\sum}}\mathrm{p}_{\boldsymbol{\mathrm{k}}} \:=\:\mathrm{2020}\:+\:\frac{\mathrm{2020}}{\mathrm{2021}}\:\boldsymbol{\beta}\:=\:\mathrm{2020}\:\blacktriangle \\ $$
Answered by mr W last updated on 09/Sep/21
1+x+x^2 +...+x^n =((1βˆ’x^(n+1) )/(1βˆ’x)) A=Ξ£_(n=1) ^(2021) ((1βˆ’x)/(1βˆ’x^(n+1) ))=Ξ£_(n=1) ^(2021) (((1/x^(n+1) )βˆ’(1/x^n ))/((1/x^(n+1) )βˆ’1)) =Ξ£_(n=1) ^(2021) ((2021^(n+1) βˆ’2021^n )/(2021^(n+1) βˆ’1)) =Ξ£_(n=1) ^(2021) [1βˆ’((2021^n βˆ’1)/(2021^(n+1) βˆ’1))] =Ξ£_(n=1) ^(2021) [1βˆ’((1βˆ’(1/(2021^n )))/(2021βˆ’(1/(2021^n ))))] =2021βˆ’(1/(2021))Ξ£_(n=1) ^(2021) [((1βˆ’(1/(2021^n )))/(1βˆ’(1/(2021^(n+1) ))))] 0<((1βˆ’(1/(2021^n )))/(1βˆ’(1/(2021^(n+1) ))))<((1βˆ’(1/(2021^n )))/(1βˆ’(1/(2021^n ))))=1 0<Ξ£_(n=1) ^(2021) ((1βˆ’(1/(2021^n )))/(1βˆ’(1/(2021^(n+1) ))))<2021Γ—1=2021 0<(1/(2021))Ξ£_(n=1) ^(2021) ((1βˆ’(1/(2021^n )))/(1βˆ’(1/(2021^(n+1) ))))<1 A=2021βˆ’(1/(2021))Ξ£_(n=1) ^(2021) [((1βˆ’(1/(2021^n )))/(1βˆ’(1/(2021^(n+1) ))))]<2021 A=2021βˆ’(1/(2021))Ξ£_(n=1) ^(2021) [((1βˆ’(1/(2021^n )))/(1βˆ’(1/(2021^(n+1) ))))]>2021βˆ’1=2020 β‡’integer part of A is 2020.
$$\mathrm{1}+{x}+{x}^{\mathrm{2}} +…+{x}^{{n}} =\frac{\mathrm{1}βˆ’{x}^{{n}+\mathrm{1}} }{\mathrm{1}βˆ’{x}} \\ $$$${A}=\underset{{n}=\mathrm{1}} {\overset{\mathrm{2021}} {\sum}}\frac{\mathrm{1}βˆ’{x}}{\mathrm{1}βˆ’{x}^{{n}+\mathrm{1}} }=\underset{{n}=\mathrm{1}} {\overset{\mathrm{2021}} {\sum}}\frac{\frac{\mathrm{1}}{{x}^{{n}+\mathrm{1}} }βˆ’\frac{\mathrm{1}}{{x}^{{n}} }}{\frac{\mathrm{1}}{{x}^{{n}+\mathrm{1}} }βˆ’\mathrm{1}} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{2021}} {\sum}}\frac{\mathrm{2021}^{{n}+\mathrm{1}} βˆ’\mathrm{2021}^{{n}} }{\mathrm{2021}^{{n}+\mathrm{1}} βˆ’\mathrm{1}} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{2021}} {\sum}}\left[\mathrm{1}βˆ’\frac{\mathrm{2021}^{{n}} βˆ’\mathrm{1}}{\mathrm{2021}^{{n}+\mathrm{1}} βˆ’\mathrm{1}}\right] \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{2021}} {\sum}}\left[\mathrm{1}βˆ’\frac{\mathrm{1}βˆ’\frac{\mathrm{1}}{\mathrm{2021}^{{n}} }}{\mathrm{2021}βˆ’\frac{\mathrm{1}}{\mathrm{2021}^{{n}} }}\right] \\ $$$$=\mathrm{2021}βˆ’\frac{\mathrm{1}}{\mathrm{2021}}\underset{{n}=\mathrm{1}} {\overset{\mathrm{2021}} {\sum}}\left[\frac{\mathrm{1}βˆ’\frac{\mathrm{1}}{\mathrm{2021}^{{n}} }}{\mathrm{1}βˆ’\frac{\mathrm{1}}{\mathrm{2021}^{{n}+\mathrm{1}} }}\right] \\ $$$$\mathrm{0}<\frac{\mathrm{1}βˆ’\frac{\mathrm{1}}{\mathrm{2021}^{{n}} }}{\mathrm{1}βˆ’\frac{\mathrm{1}}{\mathrm{2021}^{{n}+\mathrm{1}} }}<\frac{\mathrm{1}βˆ’\frac{\mathrm{1}}{\mathrm{2021}^{{n}} }}{\mathrm{1}βˆ’\frac{\mathrm{1}}{\mathrm{2021}^{{n}} }}=\mathrm{1} \\ $$$$\mathrm{0}<\underset{{n}=\mathrm{1}} {\overset{\mathrm{2021}} {\sum}}\frac{\mathrm{1}βˆ’\frac{\mathrm{1}}{\mathrm{2021}^{{n}} }}{\mathrm{1}βˆ’\frac{\mathrm{1}}{\mathrm{2021}^{{n}+\mathrm{1}} }}<\mathrm{2021}Γ—\mathrm{1}=\mathrm{2021} \\ $$$$\mathrm{0}<\frac{\mathrm{1}}{\mathrm{2021}}\underset{{n}=\mathrm{1}} {\overset{\mathrm{2021}} {\sum}}\frac{\mathrm{1}βˆ’\frac{\mathrm{1}}{\mathrm{2021}^{{n}} }}{\mathrm{1}βˆ’\frac{\mathrm{1}}{\mathrm{2021}^{{n}+\mathrm{1}} }}<\mathrm{1} \\ $$$${A}=\mathrm{2021}βˆ’\frac{\mathrm{1}}{\mathrm{2021}}\underset{{n}=\mathrm{1}} {\overset{\mathrm{2021}} {\sum}}\left[\frac{\mathrm{1}βˆ’\frac{\mathrm{1}}{\mathrm{2021}^{{n}} }}{\mathrm{1}βˆ’\frac{\mathrm{1}}{\mathrm{2021}^{{n}+\mathrm{1}} }}\right]<\mathrm{2021} \\ $$$${A}=\mathrm{2021}βˆ’\frac{\mathrm{1}}{\mathrm{2021}}\underset{{n}=\mathrm{1}} {\overset{\mathrm{2021}} {\sum}}\left[\frac{\mathrm{1}βˆ’\frac{\mathrm{1}}{\mathrm{2021}^{{n}} }}{\mathrm{1}βˆ’\frac{\mathrm{1}}{\mathrm{2021}^{{n}+\mathrm{1}} }}\right]>\mathrm{2021}βˆ’\mathrm{1}=\mathrm{2020} \\ $$$$ \\ $$$$\Rightarrow{integer}\:{part}\:{of}\:{A}\:{is}\:\mathrm{2020}. \\ $$
Commented by mathdanisur last updated on 09/Sep/21
Very nice solution, thank you Ser
$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{solution},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{Ser} \\ $$

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