Предмет: Математика,
автор: Наталинас
Помогите решить номер 23 а) и в)
Приложения:
Ответы
Автор ответа:
0
a)
![frac{A_n^7-A_n^9}{A_n^8} = frac{ frac{n!}{(n-7)!}- frac{n!}{(n-9)!} }{ frac{n!}{(n-8)!} } = frac{(n-8)!(1-(n-8)(n-7))}{(n-7)!} = frac{1-n^2+15n-56}{n-7} =\
= frac{n^2-15n+55}{7-n}](https://tex.z-dn.net/?f=+frac%7BA_n%5E7-A_n%5E9%7D%7BA_n%5E8%7D+%3D+frac%7B+frac%7Bn%21%7D%7B%28n-7%29%21%7D-+frac%7Bn%21%7D%7B%28n-9%29%21%7D++%7D%7B+frac%7Bn%21%7D%7B%28n-8%29%21%7D+%7D+%3D+frac%7B%28n-8%29%21%281-%28n-8%29%28n-7%29%29%7D%7B%28n-7%29%21%7D+%3D+frac%7B1-n%5E2%2B15n-56%7D%7Bn-7%7D+%3D%5C%0A%3D+frac%7Bn%5E2-15n%2B55%7D%7B7-n%7D+ "frac{A_n^7-A_n^9}{A_n^8} = frac{ frac{n!}{(n-7)!}- frac{n!}{(n-9)!} }{ frac{n!}{(n-8)!} } = frac{(n-8)!(1-(n-8)(n-7))}{(n-7)!} = frac{1-n^2+15n-56}{n-7} =\
= frac{n^2-15n+55}{7-n}")
в)
![frac{A_n^3*P_n+2P_{n+1}}{P_{n+1}} = frac{ frac{n!}{(n-3)!}*n!+2(n+1)! }{(n+1)!} = frac{(n-2)(n-1)n+2(n+1)}{n+1} =\
= frac{n^3-3n^2+2n+2n+2}{n+1} = frac{n^3-3n^2+4n+2}{n+1}](https://tex.z-dn.net/?f=+frac%7BA_n%5E3%2AP_n%2B2P_%7Bn%2B1%7D%7D%7BP_%7Bn%2B1%7D%7D+%3D+frac%7B+frac%7Bn%21%7D%7B%28n-3%29%21%7D%2An%21%2B2%28n%2B1%29%21+%7D%7B%28n%2B1%29%21%7D+%3D+frac%7B%28n-2%29%28n-1%29n%2B2%28n%2B1%29%7D%7Bn%2B1%7D+%3D%5C%0A%3D+frac%7Bn%5E3-3n%5E2%2B2n%2B2n%2B2%7D%7Bn%2B1%7D+%3D+frac%7Bn%5E3-3n%5E2%2B4n%2B2%7D%7Bn%2B1%7D+ "frac{A_n^3*P_n+2P_{n+1}}{P_{n+1}} = frac{ frac{n!}{(n-3)!}*n!+2(n+1)! }{(n+1)!} = frac{(n-2)(n-1)n+2(n+1)}{n+1} =\
= frac{n^3-3n^2+2n+2n+2}{n+1} = frac{n^3-3n^2+4n+2}{n+1}")
в)
Автор ответа:
0
Спасибо большое.
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