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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

$$\frac{(m+1)!(1+3+5+...+2m+3}{2m(m+2)(1+2+3+...+m+1)}$$$$=\frac{(m+1)!\frac{m+2}{2}\times 2\{1+(m+2-1)\}}{2m(m+2)\frac{(m+1)(m+2)}{2}}$$$$=\frac{(m+1)!{(m+2)}^2}{m{(m+2)}^2(m+1)}$$$$=\frac{m(m+1)\times (m-1)!}{m(m+1)}$$$$=(m-1)!$$

Answered by MathematicalUser2357 last updated on 14/Apr/24

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