Question Number 153629 by mathdanisur last updated on 08/Sep/21
Commented by benhamimed last updated on 09/Sep/21
![1+x+...+x^n =((1−x^n )/(1−x))=(((t^n −1)/t^n )/((t−1)/t))=(((t^n −1))/(t^(n−1) (t−1))) tq x=(1/t) si x=(1/(2021)) ;t=2021 A=(1/(2021−1))[((2021−1)/(2021^0 ))+((2021^2 −1)/(2021^1 ))+....+((2021^(2021) −1)/(2021^(2020) ))] A=(1/(2020))[(((2021^(2021) −2021^(2020) )+(2021^(2021) −2021^(2019) )+...+(2021^(2021) −2021^0 ))/(2021^(2020) ))]](https://www.tinkutara.com/question/Q153666.png)
$$\mathrm{1}+{x}+…+{x}^{{n}} =\frac{\mathrm{1}−{x}^{{n}} }{\mathrm{1}−{x}}=\frac{\frac{{t}^{{n}} −\mathrm{1}}{{t}^{{n}} }}{\frac{{t}−\mathrm{1}}{{t}}}=\frac{\left({t}^{{n}} −\mathrm{1}\right)}{{t}^{{n}−\mathrm{1}} \left({t}−\mathrm{1}\right)}\:\:\:\:{tq}\:{x}=\frac{\mathrm{1}}{{t}} \\ $$$${si}\:{x}=\frac{\mathrm{1}}{\mathrm{2021}}\:\:;{t}=\mathrm{2021} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2021}−\mathrm{1}}\left[\frac{\mathrm{2021}−\mathrm{1}}{\mathrm{2021}^{\mathrm{0}} }+\frac{\mathrm{2021}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2021}^{\mathrm{1}} }+….+\frac{\mathrm{2021}^{\mathrm{2021}} −\mathrm{1}}{\mathrm{2021}^{\mathrm{2020}} }\right] \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2020}}\left[\frac{\left(\mathrm{2021}^{\mathrm{2021}} −\mathrm{2021}^{\mathrm{2020}} \right)+\left(\mathrm{2021}^{\mathrm{2021}} −\mathrm{2021}^{\mathrm{2019}} \right)+…+\left(\mathrm{2021}^{\mathrm{2021}} −\mathrm{2021}^{\mathrm{0}} \right)}{\mathrm{2021}^{\mathrm{2020}} }\right] \\ $$
Commented by benhamimed last updated on 09/Sep/21
$${A}=\frac{\mathrm{1}}{\mathrm{2020}}\left(\frac{\mathrm{2021}×\mathrm{2021}^{\mathrm{2021}} −\left(\mathrm{2021}^{\mathrm{0}} +…..+\mathrm{2021}^{\mathrm{2020}} \right)}{\mathrm{2021}^{\mathrm{2020}} }\right) \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2020}}\left(\mathrm{2021}^{\mathrm{2}} −\frac{\mathrm{2021}^{\mathrm{2021}} −\mathrm{1}}{\mathrm{2020}×\mathrm{2021}^{\mathrm{2020}} }\right) \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2020}}\left(\mathrm{2021}^{\mathrm{2}} −\frac{\mathrm{2021}}{\mathrm{2020}}+\frac{\mathrm{1}}{\mathrm{2020}×\mathrm{2021}^{\mathrm{2020}} }\right) \\ $$$${E}\left({A}\right)=\mathrm{2021} \\ $$
Commented by mathdanisur last updated on 09/Sep/21
$$\mathrm{No}\:\mathrm{ser},\:=\:\mathrm{2020} \\ $$