Question Number 209718 by Ismoiljon_008 last updated on 19/Jul/24
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Answered by mr W last updated on 19/Jul/24
![a_n =(1/(n!×(n+2)))=(1/((n+1)!))−(1/((n+2)!)) Σ_(n=1) ^(2023) a_n =((1/(2!))−(1/(3!)))+((1/(3!))−(1/(4!)))+...+((1/(2024!))−(1/(2025!)) =(1/2)−(1/(2025!)) ⇒C)](https://www.tinkutara.com/question/Q209724.png)
$${a}_{{n}} =\frac{\mathrm{1}}{{n}!×\left({n}+\mathrm{2}\right)}=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!}−\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)!} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}{a}_{{n}} =\left(\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}!}−\frac{\mathrm{1}}{\mathrm{4}!}\right)+…+\left(\frac{\mathrm{1}}{\mathrm{2024}!}−\frac{\mathrm{1}}{\mathrm{2025}!}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2025}!}\:\:\Rightarrow{C}\right) \\ $$
Commented by Ismoiljon_008 last updated on 19/Jul/24
![thank you Mr W I appreciate you](https://www.tinkutara.com/question/Q209730.png)
$$\:\:\:{thank}\:{you}\:{Mr}\:{W} \\ $$$$\:\:\:{I}\:{appreciate}\:{you} \\ $$