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 175230 by infinityaction last updated on 24/Aug/22

Answered by Ar Brandon last updated on 24/Aug/22

$$S=\underset{n=3}{\sum ^{\mathrm{\infty }}}\frac{1}{\stackrel{n}{}{C}_3}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{n!\times 3!}{(n+3)!}$$$$=6\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(n+3)(n+2)(n+1)}$$$$=6\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{2(n+1)}-\frac{1}{n+2}+\frac{1}{2(n+3)})$$$$=6(\frac{1}{2}(T+1)-T+\frac{1}{2}(T-\frac{1}{2}))$$$$T=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n+2}\Rightarrow \underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n+1}=T+1,$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n+3}=T-\frac{1}{2};$$$$=6(\frac{1}{2}-\frac{1}{4})=6\left(\frac{1}{4}\right)=\frac{3}{2}★$$

Commented by infinityaction last updated on 24/Aug/22

$$sirexplainredterm$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{n!\times 3!}{(n+3)!}=6\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{(n+1)!-n!}{(n+3)!}$$$$$$

Commented by infinityaction last updated on 24/Aug/22

$$okeysirthankyou$$

Commented by Ar Brandon last updated on 24/Aug/22

Sorry, mistake. Edited!

Commented by Tawa11 last updated on 25/Aug/22

$$\mathrm{Great}\mathrm{sir}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com