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Question Number 65992 by naka3546 last updated on 07/Aug/19

x = (1/(3∙1!)) + (1/(4∙2!)) + (1/(5∙3!)) + … + (1/(1002∙1000!)) x∙1000! = ?

$${x}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{3}\centerdot\mathrm{1}!}\:+\:\frac{\mathrm{1}}{\mathrm{4}\centerdot\mathrm{2}!}\:+\:\frac{\mathrm{1}}{\mathrm{5}\centerdot\mathrm{3}!}\:+\:\ldots\:+\:\frac{\mathrm{1}}{\mathrm{1002}\centerdot\mathrm{1000}!} \\ $$$${x}\centerdot\mathrm{1000}!\:\:=\:\:? \\ $$

Commented by Prithwish sen last updated on 08/Aug/19

last term (n/((n+1)!)) where n= 1001 ∴ [((n+1−1)/((n+1)!))] = (1/(n!)) − (1/((n+1)!)) ∴ x = (2/(3!))+(3/(4!))+...+((1001)/(1002! )) = (1/(2!))−(1/(3!))+(1/(3!))−(1/(4!))+...+(1/(1001!)) −(1/(1002!)) =(1/(2!))−(1/(1002!))

$$\mathrm{last}\:\mathrm{term}\:\frac{\mathrm{n}}{\left(\mathrm{n}+\mathrm{1}\right)!}\:\:\mathrm{where}\:\mathrm{n}=\:\mathrm{1001} \\ $$$$\therefore\:\left[\frac{\mathrm{n}+\mathrm{1}−\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)!}\right]\:=\:\frac{\mathrm{1}}{\mathrm{n}!}\:−\:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)!} \\ $$$$\therefore\:\mathrm{x}\:=\:\frac{\mathrm{2}}{\mathrm{3}!}+\frac{\mathrm{3}}{\mathrm{4}!}+...+\frac{\mathrm{1001}}{\mathrm{1002}!\:}\:=\:\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{3}!}−\frac{\mathrm{1}}{\mathrm{4}!}+...+\frac{\mathrm{1}}{\mathrm{1001}!}\:−\frac{\mathrm{1}}{\mathrm{1002}!} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{1002}!} \\ $$

Answered by $@ty@m123 last updated on 07/Aug/19

x=(2/(3!))+(3/(4!))+....+((1001)/(1002!)) x=(3/(3!))−(1/(3!))+(4/(4!))−(1/(4!))+....+((1002)/(1002!))−(1/(1002!)) x=(1/(2!))−(1/(3!))+(1/(3!))−(1/(4!))+......+(1/(1001!))−(1/(1002!)) x=(1/(2!))−(1/(1002!)) x.1000!=((1000!)/2)−((1000!)/(1002!)) x.1000!=((1000!)/2)−(1/(1002.1001)) x.1000!=((501.1001!−1)/(1002.1001))

$${x}=\frac{\mathrm{2}}{\mathrm{3}!}+\frac{\mathrm{3}}{\mathrm{4}!}+....+\frac{\mathrm{1001}}{\mathrm{1002}!} \\ $$$${x}=\frac{\mathrm{3}}{\mathrm{3}!}−\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{4}}{\mathrm{4}!}−\frac{\mathrm{1}}{\mathrm{4}!}+....+\frac{\mathrm{1002}}{\mathrm{1002}!}−\frac{\mathrm{1}}{\mathrm{1002}!} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{3}!}−\frac{\mathrm{1}}{\mathrm{4}!}+......+\frac{\mathrm{1}}{\mathrm{1001}!}−\frac{\mathrm{1}}{\mathrm{1002}!} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{1002}!} \\ $$$${x}.\mathrm{1000}!=\frac{\mathrm{1000}!}{\mathrm{2}}−\frac{\mathrm{1000}!}{\mathrm{1002}!} \\ $$$${x}.\mathrm{1000}!=\frac{\mathrm{1000}!}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{1002}.\mathrm{1001}} \\ $$$${x}.\mathrm{1000}!=\frac{\mathrm{501}.\mathrm{1001}!−\mathrm{1}}{\mathrm{1002}.\mathrm{1001}} \\ $$$$ \\ $$

Commented by Prithwish sen last updated on 07/Aug/19

beautiful.

$$\mathrm{beautiful}. \\ $$

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