Category: Algebra

Question Number 191833 by Mingma last updated on 01/May/23 Answered by AST last updated on 01/May/23 $$a={log}_{2}900;b={log}_{3}900;c={log}_{5}900$$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}={log}_{\mathrm{900}} \mathrm{2}+{log}_{\mathrm{900}} \mathrm{3}+{log}_{\mathrm{900}} \mathrm{5}…

Question Number 60745 by Tawa1 last updated on 25/May/19 Commented by Prithwish sen last updated on 25/May/19 $$\frac{{[(\mathrm{n}+1)!]}^{\mathrm{n}}}{{(\mathrm{n}!)}^{\mathrm{n}+1}}={\left[\frac{(\mathrm{n}+1)!}{\mathrm{n}!}\right]}^{\mathrm{n}}.\frac{1}{\mathrm{n}!}=\frac{{(\mathrm{n}+1)}^{\mathrm{n}}}{\mathrm{n}!}$$$$=\frac{\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{1}}.\frac{\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}…………\frac{\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{n}}\:>\mathrm{1} \…

Question Number 191798 by Abdullahrussell last updated on 30/Apr/23 Commented by Frix last updated on 30/Apr/23 $$\mathrm{Qu}191753;\mathrm{just}\mathrm{add}-3+24=21$$ Commented by BaliramKumar last updated on…

Question Number 191790 by mathlove last updated on 30/Apr/23 $$provethat$$$$\frac{2x-4}{2\cdot 3\cdot 4}+\frac{3x-5}{3\cdot 4\cdot 5}+\frac{4x-6}{4\cdot 5\cdot 6}+\dots ..+\frac{100x-102}{100\cdot 101\cdot 102}=\frac{103}{102}$$$$$$ Commented by BaliramKumar last updated on 01/May/23 $$\mathrm{if}\:\mathrm{x}=\mathrm{1}\:\mathrm{then}\:\mathrm{all}\:\mathrm{term}\:\left(−\mathrm{ve}\right)\:\mathrm{but}\:\frac{\mathrm{103}}{\mathrm{102}}\:\mathrm{is}\:\left(+\mathrm{ve}\right)\: \…

Question Number 191775 by Mingma last updated on 30/Apr/23 Answered by AST last updated on 30/Apr/23 $$\stackrel{\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}}{ABCDEF}=10000\stackrel{\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}}{ABCD}+\stackrel{\mathrm{\_}\mathrm{\_}\mathrm{\_}}{EF}$$$$\mathrm{17}\left[\mathrm{10000}\overset{\_\_\_} {{AB}}+\overset{\_\_\_\_\_\_\_} {{CDEF}}\right]=\mathrm{12}\left[\mathrm{100}\overset{\_\_\_\_\_\_\_} {{CDEF}}+\overset{\_\_\_}…