Question Number 218854 by Rojarani last updated on 16/Apr/25 Answered by Hamada1969 last updated on 16/Apr/25 $${k}\:=\:\mathrm{15127} \\ $$$$ \\ $$ Commented by Rojarani last…
Question Number 218848 by hardmath last updated on 16/Apr/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 218855 by Lekhraj last updated on 16/Apr/25 $$\sqrt{\mathrm{3}}{x}^{\mathrm{2}} =\sqrt{\left(\mathrm{64}−{x}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −\mathrm{4}\right)}+\sqrt{\left(\mathrm{81}−{x}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −\mathrm{1}\right)}+\sqrt{\left(\mathrm{49}−{x}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −\mathrm{1}\right)}\boldsymbol{{solve}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 218849 by hardmath last updated on 16/Apr/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 218850 by hardmath last updated on 16/Apr/25 Answered by MrGaster last updated on 17/Apr/25 $${T}_{{n}} =\frac{\mathrm{1}}{\mathrm{ln}\left(\underset{{r}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{{r}^{\mathrm{2}} }{\mathrm{1}+{r}^{\mathrm{2}} }\right)}=\frac{\mathrm{1}}{\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{ln}\left(\frac{{r}^{\mathrm{2}} }{\mathrm{1}+{r}^{\mathrm{2}}…
Question Number 218820 by universe last updated on 16/Apr/25 Answered by mr W last updated on 17/Apr/25 $${a}_{{n}} ={a}+{b}−\frac{{ab}}{{a}_{{n}−\mathrm{1}} } \\ $$$${let}\:{a}_{{n}} =\frac{{t}_{{n}} }{{t}_{{n}−\mathrm{1}} }…
Question Number 218749 by hardmath last updated on 15/Apr/25 Answered by MrGaster last updated on 17/Apr/25 $${T}_{{n}} =\frac{\mathrm{1}}{\mathrm{ln}\left(\underset{{r}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{{r}^{\mathrm{2}} }{\mathrm{1}+{r}^{\mathrm{2}} }\right)}=\frac{\mathrm{1}}{\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{ln}\left(\frac{{r}^{\mathrm{2}} }{\mathrm{1}+{r}^{\mathrm{2}}…
Question Number 218651 by hardmath last updated on 13/Apr/25 Answered by vnm last updated on 14/Apr/25 $$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{4}\centerdot\frac{\mathrm{1}}{\mathrm{4}}\right)+ \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{8}}−\mathrm{4}\centerdot\frac{\mathrm{1}}{\mathrm{8}}\right)+ \\ $$$$…+ \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{4}{n}−\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}{n}−\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}{n}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{4}{n}}−\mathrm{4}\centerdot\frac{\mathrm{1}}{\mathrm{4}{n}}\right)= \\ $$$$\underset{{k}=\mathrm{1}}…
Question Number 218636 by hardmath last updated on 13/Apr/25 Answered by vnm last updated on 14/Apr/25 $${let} \\ $$$$\:{a}_{{k}} =\frac{\mathrm{1}}{\mathrm{3}{k}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{3}{k}−\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{3}{k}} \\ $$$${b}_{{k}} =\frac{\mathrm{4}{k}}{\mathrm{4}{k}+\mathrm{1}}−\frac{\mathrm{4}\left({k}−\mathrm{1}\right)}{\mathrm{4}\left({k}−\mathrm{1}\right)+\mathrm{1}} \\ $$$${if}\:\:\forall{k}\geqslant\mathrm{1}\:{a}_{{k}}…
Question Number 218603 by hardmath last updated on 12/Apr/25 Commented by mr W last updated on 13/Apr/25 $${seems}\:{true} \\ $$ Commented by mr W last…