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