Выпишите выражения, равные произведениям: (n-1)(n-2)(n-3)(n-4)(n-5) (1-n)(n-2)(n-3)(n-4)(n-5)(n-1)(2-n)(n-3)(n-4)(n-5)(1-n)(2-n)(3-n)(4-n)(5-n)

Ответ:

polulera2005

30.09.2020 21:33

1) Полагаю, что n ∈ Z

$if \ n 5, \ 0! = 1, \ n! = 1*2*...*n\\\\ (n-1)*(n-2)*(n-3)*(n-4)*(n-5) =\\ (n-1)!/(n-6)!\\\\ (1-n)*(n-2)*(n-3)*(n-4)*(n-5) =\\ -(n-1)*(n-2)*(n-3)*(n-4)*(n-5) =\\-(n-1)!/(n-6)!\\\\ (n-1)*(2-n)*(n-3)*(n-4)*(n-5) =\\ -(n-1)*(n-2)*(n-3)*(n-4)*(n-5) =\\ -(n-1)!/(n-6)!\\\\ (1-n)*(2-n)*(3-n)*(4-n)*(5-n) =\\ (-1)^5(n-1)*(n-2)*(n-3)*(n-4)*(n-5) =\\ -(n-1)!/(n-6)!\\\\$

$if \ n = 1, 2, 3, 4 \ or \ 5\\\\ 0\\\\ if \ n < 0, \ -n = m, \ m 0\\\\ (n-1)*(n-2)*(n-3)*(n-4)*(n-5) =\\ (-m-1)*(-m-2)*(-m-3)*(-m-4)*(-m-5) =\\ (-1)^5(m+1)*(m+2)*(m+3)*(m+4)*(m+5) =\\ -(m+5)!/m!\\\\$

$(1-n)*(n-2)*(n-3)*(n-4)*(n-5) =\\ -(n-1)*(n-2)*(n-3)*(n-4)*(n-5) =\\ (-1)^6(m+1)*(m+2)*(m+3)*(m+4)*(m+5) =\\ (m+5)!/m!\\\\ (n-1)*(2-n)*(n-3)*(n-4)*(n-5) =\\ (-1)^6(m+1)*(m+2)*(m+3)*(m+4)*(m+5) =\\ (m+5)!/m!\\\\ (1-n)*(2-n)*(3-n)*(4-n)*(5-n) =\\ (-1)^5(n-1)*(n-2)*(n-3)*(n-4)*(n-5)=\\ (-1)^{10}(m+1)*(m+2)*(m+3)*(m+4)*(m+5) =\\ (m+5)!/m!$

$2) \ (n-1)*(n-2)*(n-3)*(n-4)*(n-5)=\\ (n^2-3n+2)(n^2-7n+12)(n-5) =\\ (n^2-3n+2)(n^3-12 n^2+47 n-60) =\\ n^5 - 12n^4 +47n^3-60n^2-3n^4+36n^3-141n^2+\\180n+2n^3-24n^2+94n-120 =\\ n^5 - 15n^4 + 85n^3 - 225n^2 + 274n - 120\\\\ (1-n)*(n-2)*(n-3)*(n-4)*(n-5) =\\ -(n-1)*(n-2)*(n-3)*(n-4)*(n-5) =\\-n^5+15n^4 - 85n^3 + 225n^2 - 274n +120\\\\$

$(1-n)*(2-n)*(n-3)*(n-4)*(n-5) =\\ -(n-1)*(n-2)*(n-3)*(n-4)*(n-5) =\\-n^5+15n^4 - 85n^3 + 225n^2 - 274n +120\\\\ (1-n)*(2-n)*(3-n)*(4-n)*(5-n) =\\ (-1)^5(n-1)*(n-2)*(n-3)*(n-4)*(n-5) =\\-(n-1)*(n-2)*(n-3)*(n-4)*(n-5) =\\-n^5+15n^4 - 85n^3 + 225n^2 - 274n +120$