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Question Number 202161 by Calculusboy last updated on 22/Dec/23

Answered by som(math1967) last updated on 22/Dec/23

(((m+1)!(1+3+5+...+2m+3)/(2m(m+2)(1+2+3+...+m+1))) =(((m+1)!((m+2)/2)×2{1+(m+2−1)})/(2m(m+2)(((m+1)(m+2))/2))) =(((m+1)!(m+2)^2 )/(m(m+2)^2 (m+1))) =((m(m+1)×(m−1)!)/(m(m+1))) =(m−1)!

$$\frac{\left({m}+\mathrm{1}\right)!\left(\mathrm{1}+\mathrm{3}+\mathrm{5}+...+\mathrm{2}{m}+\mathrm{3}\right.}{\mathrm{2}{m}\left({m}+\mathrm{2}\right)\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+...+{m}+\mathrm{1}\right)} \\ $$$$=\frac{\left({m}+\mathrm{1}\right)!\frac{{m}+\mathrm{2}}{\mathrm{2}}×\mathrm{2}\left\{\mathrm{1}+\left({m}+\mathrm{2}−\mathrm{1}\right)\right\}}{\mathrm{2}{m}\left({m}+\mathrm{2}\right)\frac{\left({m}+\mathrm{1}\right)\left({m}+\mathrm{2}\right)}{\mathrm{2}}} \\ $$$$=\frac{\left({m}+\mathrm{1}\right)!\left({m}+\mathrm{2}\right)^{\mathrm{2}} }{{m}\left({m}+\mathrm{2}\right)^{\mathrm{2}} \left({m}+\mathrm{1}\right)} \\ $$$$=\frac{{m}\left({m}+\mathrm{1}\right)×\left({m}−\mathrm{1}\right)!}{{m}\left({m}+\mathrm{1}\right)} \\ $$$$=\left({m}−\mathrm{1}\right)!\: \\ $$

Answered by MathematicalUser2357 last updated on 14/Apr/24

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