Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 214310 by malwan last updated on 04/Dec/24

$$\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}$$

Answered by mr W last updated on 05/Dec/24

$$\frac{n-1}{n!}=\frac{n}{n!}-\frac{1}{n!}=\frac{1}{(n-1)!}-\frac{1}{n!}$$$$$$$$sum=(\frac{1}{1!}-\frac{1}{2!})+(\frac{1}{2!}-\frac{1}{3!})+(\frac{1}{3!}-\frac{1}{4!})+...+(\frac{1}{99!}-\frac{1}{100!})$$$$=1-\frac{1}{100!}$$

Commented by malwan last updated on 05/Dec/24

$$thankyousomuchsir$$

Answered by Frix last updated on 04/Dec/24

$$S_n=\underset{k=1}{\sum ^n}\frac{k}{(k+1)!}=1-\frac{1}{(n+1)!}$$$$S_{99}=1-\frac{1}{100!}$$

Commented by malwan last updated on 05/Dec/24

$$thankyousir$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com