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 152030 by mathdanisur last updated on 25/Aug/21

Commented by mathdanisur last updated on 26/Aug/21

$$\mathbf{S}\mathrm{er},{\mathrm{\Omega }}_n+{\mathrm{\Omega }}_{n+2}+...+{\mathrm{\Omega }}_{n\times 6}=1$$

Commented by ghimisi last updated on 25/Aug/21

$${\mathrm{\Omega }}_p+{\mathrm{\Omega }}_{p+2}=\frac{1}{p+1}$$

Answered by Kamel last updated on 25/Aug/21

$${\mathrm{\Omega }}_n+{\mathrm{\Omega }}_{n+2}={\int }_0^{\frac{\pi }{4}}{tan}^n\left(x\right)(1+{tan}^2\left(x\right))dx={\int }_0^1{t}^{n}dt$$$$=\frac{1}{n+1}$$$$$$

Commented by mathdanisur last updated on 26/Aug/21

$$\mathrm{Ser},{\mathrm{\Omega }}_n+{\mathrm{\Omega }}_{n+2}+{\mathrm{\Omega }}_{n+3}+...+{\mathrm{\Omega }}_{n+6}=1$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com