Question Number 209718 by Ismoiljon_008 last updated on 19/Jul/24
Answered by mr W last updated on 19/Jul/24
$${a}_{{n}} =\frac{\mathrm{1}}{{n}!×\left({n}+\mathrm{2}\right)}=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!}−\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)!} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}{a}_{{n}} =\left(\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}!}−\frac{\mathrm{1}}{\mathrm{4}!}\right)+...+\left(\frac{\mathrm{1}}{\mathrm{2024}!}−\frac{\mathrm{1}}{\mathrm{2025}!}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2025}!}\:\:\Rightarrow{C}\right) \\ $$
Commented by Ismoiljon_008 last updated on 19/Jul/24
$$\:\:\:{thank}\:{you}\:{Mr}\:{W} \\ $$$$\:\:\:{I}\:{appreciate}\:{you} \\ $$