Question Number 25170 by NECx last updated on 05/Dec/17 $${prove}\:{that}\:{n}^{\mathrm{2}} >{n}−\mathrm{5}\:{for}\:{integral}\: \\ $$$${n}\geqslant\mathrm{3}\: \\ $$ Commented by mrW1 last updated on 06/Dec/17 $${maybe}\:{the}\:{question}\:{is} \\ $$$${n}^{\mathrm{2}}…
Question Number 25171 by NECx last updated on 05/Dec/17 $$\mathrm{100}{n}>{n}^{\mathrm{2}} \:{for}\:{integral}\:{n}>\mathrm{100} \\ $$$$ \\ $$ Answered by ajfour last updated on 05/Dec/17 $${n}\:>\:\mathrm{100} \\ $$$${n}={n}…
Question Number 156213 by MathSh last updated on 09/Oct/21 $$\Omega\:=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\mathrm{log}^{\mathrm{2}} \:\left(\frac{\Gamma\left(\mathrm{x}+\mathrm{1}\right)}{\mathrm{x}}\right)\:\mathrm{dx}\:=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 156214 by MathSh last updated on 09/Oct/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 25122 by math solver last updated on 04/Dec/17 $$\:\mathrm{3}_{{C}_{\mathrm{1}} } \:+\:\mathrm{4}_{{C}_{\mathrm{2}} } \:+\:\mathrm{5}_{{C}_{\mathrm{3}} } \:+………..+\:\mathrm{49}_{{C}_{\mathrm{47}} } \:=\:? \\ $$$${where}\:{n}_{{C}_{{r}} } \:=\:\frac{{n}!}{{r}!×\left({n}−{r}\right)!}\:. \\ $$…
Question Number 90661 by Cynosure last updated on 25/Apr/20 $${show}\:{that}\:\left({n}^{\mathrm{4}} −{n}^{\mathrm{2}} \right)\:{is}\:{divisible}\:{by}\:\mathrm{12} \\ $$ Answered by MJS last updated on 25/Apr/20 $${n}^{\mathrm{4}} −{n}^{\mathrm{2}} ={n}^{\mathrm{2}} \left({n}^{\mathrm{2}}…
Question Number 25113 by Mr easy last updated on 04/Dec/17 Commented by prakash jain last updated on 05/Dec/17 $$\left(\mathrm{log}_{\mathrm{2}} {x}\right)^{\mathrm{3}} +\left(\mathrm{log}_{\mathrm{2}} {y}\right)^{\mathrm{3}} \\ $$$$=\left(\mathrm{log}_{\mathrm{2}} {x}+\mathrm{log}_{\mathrm{2}}…
Question Number 25114 by Mr easy last updated on 04/Dec/17 Commented by moxhix last updated on 04/Dec/17 $${x}=\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+{x}}}}>\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{0}}} \\ $$$$\therefore{x}>\sqrt{\mathrm{2}} \\ $$$${put}\:\phi:\:\:\phi=\sqrt{\mathrm{1}+\phi}\:\:\left(\phi>\mathrm{1}\right) \\ $$$$\phi^{\mathrm{2}} −\phi−\mathrm{1}=\mathrm{0}…
Question Number 25112 by Mr easy last updated on 04/Dec/17 Commented by prakash jain last updated on 04/Dec/17 $$\left({a}+{b}\right)^{\mathrm{2}{n}} =\underset{{i}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}\:^{\mathrm{2}{n}} {C}_{{i}} {a}^{{i}} {b}^{\mathrm{2}{n}−{i}}…
Question Number 156180 by SANOGO last updated on 08/Oct/21 Terms of Service Privacy Policy Contact: info@tinkutara.com