Category: Algebra

Question Number 126950 by AST last updated on 05/Dec/22 Answered by floor(10²Eta[1]) last updated on 25/Dec/20 $$\mathrm{A}.\mathrm{1}+\mathrm{2}^{\mathrm{2}^{\mathrm{n}} } +\mathrm{2}^{\mathrm{2}^{\mathrm{n}+\mathrm{1}} } \equiv\mathrm{1}+\left(−\mathrm{1}\right)^{\mathrm{2}^{\mathrm{n}} } +\left(−\mathrm{1}\right)^{\mathrm{2}^{\mathrm{n}+\mathrm{1}} } \equiv\mathrm{1}+\mathrm{1}+\mathrm{1}\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{3}\right)…

Question Number 192481 by Tomal last updated on 19/May/23 Answered by aleks041103 last updated on 19/May/23 $$\frac{1}{2}=\frac{1}{1+x}(1+{\left(\frac{1}{1+x}\right)}^2+{\left(\frac{1}{1+x}\right)}^4+\dots )$$$$if\mid \frac{1}{1+x}\mid <1,i.e.\mid 1+x\mid >1$$$$\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{1}+{x}}\:\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }}=\frac{\mathrm{1}+{x}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} −\mathrm{1}}…

Question Number 126935 by Raxreedoroid last updated on 25/Dec/20 $$Proveorgiveacounterexample:$$$${(a+1)}^{n-1}=\underset{k=1}{\sum ^n}\left(\underset{k-1}{\stackrel{n-1}{}}\right)a^{k-1}$$$$\left(\underset{r}{\stackrel{n}{}}\right)=\frac{n!}{r!(n-r)!}$$ Terms of…

Question Number 126931 by arash sharifi last updated on 25/Dec/20 $$y={(1-x)}^{cosx}$$ Answered by mahdipoor last updated on 25/Dec/20 $$salamarashjan,soalatrokhamelbenevis$$ Terms of…

Question Number 192440 by cortano12 last updated on 18/May/23 $$\mathrm{Given}\{\begin{array}{l}\mathrm{A}=\frac{{({\mathrm{p}}^2+{\mathrm{q}}^2+{\mathrm{r}}^2)}^2}{{\left(\mathrm{pq}\right)}^2+{\left(\mathrm{pr}\right)}^2+{\left(\mathrm{qr}\right)}^2}\\ \mathrm{B}=\frac{{\mathrm{q}}^2-\mathrm{pr}}{{\mathrm{p}}^2+{\mathrm{q}}^2+{\mathrm{r}}^2}\end{array}$$$$\:\mathrm{If}\:\mathrm{p}+\mathrm{q}+\mathrm{r}=\mathrm{0}\:\mathrm{then}\:\mathrm{A}^{\mathrm{2}} −\mathrm{4B}=? \…