Ensino Médio ⇒ Fatoração-FME Combinações e Probabilidades Tópico resolvido

[luyudi]

Expressar em forma fatorada:
a)1x3x5x...x(2n-1)
Por favor quem puder me responder faça passo a passo,pois eu realmente não entendi como chegar na resposta.

Obs.: x=vezes

Resposta

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(2n)!/n!2^n

[csmarcelo]

[tex3]\small1\times3\times5\times7\times...\times(2n-1)=\frac{{\color{red}1}\times2\times{\color{red}3}\times4\times{\color{red}5}\times6\times{\color{red}7}\times...\times(2n-2)\times{\color{red}(2n-1)}}{\underbrace{2\times4\times6\times...\times(2n-2)}_{n-1\text{ fatores pares}}}=\frac{1\times2\times3\times4\times5\times6\times7\times...\times(2n-1)}{2^{n-1}\times1\times2\times3\times...\times(n-1)}=\frac{(2n-1)!}{(n-1)!\cdot2^{n-1}}=\frac{\frac{(2n)!}{2n}}{\frac{n!}{n}\cdot\frac{2^n}{2}}=\frac{(2n)!}{n!\cdot2^n}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\small1\times3\times5\times7\times...\times(2n-1)=\frac{{\color{red}1}\times2\times{\color{red}3}\times4\times{\color{red}5}\times6\times{\color{red}7}\times...\times(2n-2)\times{\color{red}(2n-1)}}{\underbrace{2\times4\times6\times...\times(2n-2)}_{n-1\text{ fatores pares}}}=\frac{1\times2\times3\times4\times5\times6\times7\times...\times(2n-1)}{2^{n-1}\times1\times2\times3\times...\times(n-1)}=\frac{(2n-1)!}{(n-1)!\cdot2^{n-1}}=\frac{\frac{(2n)!}{2n}}{\frac{n!}{n}\cdot\frac{2^n}{2}}=\frac{(2n)!}{n!\cdot2^n}[/tex3]

[luyudi]

[tex3]\small1\times3\times5\times7\times...\times(2n-1)=\frac{{\color{red}1}\times2\times{\color{red}3}\times4\times{\color{red}5}\times6\times{\color{red}7}\times...\times(2n-2)\times{\color{red}(2n-1)}}{\underbrace{2\times4\times6\times...\times(2n-2)}_{n-1\text{ fatores pares}}}=\frac{1\times2\times3\times4\times5\times6\times7\times...\times(2n-1)}{2^{n-1}\times1\times2\times3\times...\times(n-1)}=\frac{(2n-1)!}{(n-1)!\cdot2^{n-1}}=\frac{\frac{(2n)!}{2n}}{\frac{n!}{n}\cdot\frac{2^n}{2}}=\frac{(2n)!}{n!\cdot2^n}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\small1\times3\times5\times7\times...\times(2n-1)=\frac{{\color{red}1}\times2\times{\color{red}3}\times4\times{\color{red}5}\times6\times{\color{red}7}\times...\times(2n-2)\times{\color{red}(2n-1)}}{\underbrace{2\times4\times6\times...\times(2n-2)}_{n-1\text{ fatores pares}}}=\frac{1\times2\times3\times4\times5\times6\times7\times...\times(2n-1)}{2^{n-1}\times1\times2\times3\times...\times(n-1)}=\frac{(2n-1)!}{(n-1)!\cdot2^{n-1}}=\frac{\frac{(2n)!}{2n}}{\frac{n!}{n}\cdot\frac{2^n}{2}}=\frac{(2n)!}{n!\cdot2^n}[/tex3]

Muito obrigado,cara.Tinha realmente que ter uns bizus que eu não sabia,estou muito feliz de ter entendido facilmente.Obrigado!