Question Number 202161 by Calculusboy last updated on 22/Dec/23
Answered by som(math1967) last updated on 22/Dec/23
$$\frac{\left({m}+\mathrm{1}\right)!\left(\mathrm{1}+\mathrm{3}+\mathrm{5}+…+\mathrm{2}{m}+\mathrm{3}\right.}{\mathrm{2}{m}\left({m}+\mathrm{2}\right)\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{m}+\mathrm{1}\right)} \\ $$$$=\frac{\left({m}+\mathrm{1}\right)!\frac{{m}+\mathrm{2}}{\mathrm{2}}×\mathrm{2}\left\{\mathrm{1}+\left({m}+\mathrm{2}−\mathrm{1}\right)\right\}}{\mathrm{2}{m}\left({m}+\mathrm{2}\right)\frac{\left({m}+\mathrm{1}\right)\left({m}+\mathrm{2}\right)}{\mathrm{2}}} \\ $$$$=\frac{\left({m}+\mathrm{1}\right)!\left({m}+\mathrm{2}\right)^{\mathrm{2}} }{{m}\left({m}+\mathrm{2}\right)^{\mathrm{2}} \left({m}+\mathrm{1}\right)} \\ $$$$=\frac{{m}\left({m}+\mathrm{1}\right)×\left({m}−\mathrm{1}\right)!}{{m}\left({m}+\mathrm{1}\right)} \\ $$$$=\left({m}−\mathrm{1}\right)!\: \\ $$
Answered by MathematicalUser2357 last updated on 14/Apr/24
$$? \\ $$