Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 20523 by Tinkutara last updated on 27/Aug/17

$$\mathrm{Let}\mathrm{the}\mathrm{sum}\underset{n=1}{\sum ^9}\frac{1}{n(n+1)(n+2)}\mathrm{written}$$$$\mathrm{in}\mathrm{its}\mathrm{lowest}\mathrm{terms}\mathrm{be}\frac{p}{q}.\mathrm{Find}\mathrm{the}\mathrm{value}$$$$\mathrm{of}q-p.$$

Answered by ajfour last updated on 27/Aug/17

$$S=\frac{1}{2}\underset{n=1}{\sum ^9}\frac{(n+2)-n}{n(n+1)(n+2)}$$$$=\frac{1}{2}\underset{n=1}{\sum ^9}[\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}]$$$$=\frac{1}{2}[(\frac{1}{1.2}-\frac{1}{2.3})+(\frac{1}{2.3}-\frac{1}{3.4})+....$$$$....+(\frac{1}{9.10}-\frac{1}{10.11})]$$$$=\frac{1}{2}[\frac{1}{2}-\frac{1}{110}]=\frac{27}{110}$$$$\Rightarrow \frac{p}{q}=\frac{27}{110}$$$$\mathit{q}-\mathit{p}=83.$$

Commented by Tinkutara last updated on 28/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com