Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 59639 by Tawa1 last updated on 12/May/19

Commented by Tawa1 last updated on 12/May/19

$$\mathrm{Please}\mathrm{help}\mathrm{me}\mathrm{to}\mathrm{continue}.\mathrm{My}\mathrm{final}\mathrm{answer}\mathrm{is}\mathrm{wrong}.$$$$\mathrm{I}\mathrm{got}:\mathrm{S}=\frac{{2\mathrm{n}}^3+{3\mathrm{n}}^2+\mathrm{n}}{6}\mathrm{but}\mathrm{answer}\mathrm{is}\mathrm{S}=\frac{\mathrm{n}}{6}({14\mathrm{n}}^2-9\mathrm{n}+1)$$

Commented by tanmay last updated on 12/May/19

$$T_r={\left(r\times \frac{n_{c_r}}{n_{c_{r-1}}}\right)}^2={\left(\frac{r\times \frac{n!}{r!(n-r)!}}{\frac{n!}{(n-r+1)!(r-1)!}}\right)}^2$$$$={\left(\frac{r\times (r-1)!(n-r+1)!}{r!(n-r)!}\right)}^2$$$$={(n-r+1)}^2$$$$T_r={(n+1)}^2-2(n+1)r+r^2$$$$T_1={(n+1)}^2-2(n+1)\times 1+1^2$$$$T_2={(n+1)}^2-2(n+1)\times 2+2^2$$$$T_3={(n+1)}^2-2(n+1)\times 3+3^2$$$$....$$$$....$$$$T_n={(n+1)}^2-2(n+1)\times n+n^2$$$$nowaddthem$$$$S=n\times {(n+1)}^2-2(n+1)(1+2+3+..+n)+(1^2+2^2+..+n^2)$$$$S=n{(n+1)}^2-2(n+1)\times \frac{n(n+1)}{2}+\frac{n(n+1)(2n+1)}{6}$$$$S=\frac{n(2n^2+3n+1)}{6}$$$$S=\frac{2n^3+3n^2+n}{6}$$$$$$$$$$$$$$$$$$

Commented by Tawa1 last updated on 13/May/19

$$\mathrm{That}\mathrm{means}\mathrm{am}\mathrm{right}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com