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Question-212066




Question Number 212066 by Spillover last updated on 28/Sep/24
Answered by Rasheed.Sindhi last updated on 29/Sep/24
((2021!+2020!)/(2021!−2020!)) ∙ ((2020!+2019!)/(2020!−2019!)) ∙...((3!+2!)/(3!−2!)) ∙ ((2!+1!)/(2!−1!))  (({(x+1)!+x!})/({(x+1)!−x!}))=((x!{(x+1)+1})/(x!{(x+1)−1}))=((x+2)/x)  ((2020+2)/(2020))∙((2019+2)/(2019))∙...((2+2)/2)∙((1+2)/1)  (((2020+2)(2019+2)(2018+2)...(2+2)(1+2))/(2020.2019.2018....2.1))  =((2022.2021.2020....4.3)/(2020!))  =((2022.2021.2020....4.3.(2.1))/(2020!(2.1)))  =((2022.2021)/2)=2043231
$$\frac{\mathrm{2021}!+\mathrm{2020}!}{\mathrm{2021}!−\mathrm{2020}!}\:\centerdot\:\frac{\mathrm{2020}!+\mathrm{2019}!}{\mathrm{2020}!−\mathrm{2019}!}\:\centerdot…\frac{\mathrm{3}!+\mathrm{2}!}{\mathrm{3}!−\mathrm{2}!}\:\centerdot\:\frac{\mathrm{2}!+\mathrm{1}!}{\mathrm{2}!−\mathrm{1}!} \\ $$$$\frac{\left\{\left({x}+\mathrm{1}\right)!+{x}!\right\}}{\left\{\left({x}+\mathrm{1}\right)!−{x}!\right\}}=\frac{{x}!\left\{\left({x}+\mathrm{1}\right)+\mathrm{1}\right\}}{{x}!\left\{\left({x}+\mathrm{1}\right)−\mathrm{1}\right\}}=\frac{{x}+\mathrm{2}}{{x}} \\ $$$$\frac{\mathrm{2020}+\mathrm{2}}{\mathrm{2020}}\centerdot\frac{\mathrm{2019}+\mathrm{2}}{\mathrm{2019}}\centerdot…\frac{\mathrm{2}+\mathrm{2}}{\mathrm{2}}\centerdot\frac{\mathrm{1}+\mathrm{2}}{\mathrm{1}} \\ $$$$\frac{\left(\mathrm{2020}+\mathrm{2}\right)\left(\mathrm{2019}+\mathrm{2}\right)\left(\mathrm{2018}+\mathrm{2}\right)…\left(\mathrm{2}+\mathrm{2}\right)\left(\mathrm{1}+\mathrm{2}\right)}{\mathrm{2020}.\mathrm{2019}.\mathrm{2018}….\mathrm{2}.\mathrm{1}} \\ $$$$=\frac{\mathrm{2022}.\mathrm{2021}.\mathrm{2020}….\mathrm{4}.\mathrm{3}}{\mathrm{2020}!} \\ $$$$=\frac{\mathrm{2022}.\mathrm{2021}.\mathrm{2020}….\mathrm{4}.\mathrm{3}.\left(\mathrm{2}.\mathrm{1}\right)}{\mathrm{2020}!\left(\mathrm{2}.\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{2022}.\mathrm{2021}}{\mathrm{2}}=\mathrm{2043231} \\ $$

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