ВЫЧИСЛИТЬ f(n)=2n!/(2n-1)! f(4) - ? f(n)=(2n^2−3)!/(n^2−1)!9! f(3)-? f(n)=(n^2−2n−3)!/(3n−1)!(n+2)! f(5)-?. Created by LadyHouston. algebra-ru

(3n−1)!(n+2)! f(5)-?...

LadyHouston @LadyHouston

August 2022 1 5 Report

ВЫЧИСЛИТЬ
f(n)=2n!/(2n-1)! f(4) - ?

f(n)=(2n^2−3)!/(n^2−1)!9! f(3)-?

f(n)=(n^2−2n−3)!/(3n−1)!(n+2)!
f(5)-?

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NNNLLL54

Verified answer

$1)\ \ f(n)=\dfrac{2n!}{(2n-1)!} \ \ ,\\\\\\f(4)=\dfrac{2\cdot 4!}{(2\cdot 4-1)!}=\dfrac{2\cdot 4!}{7!}=\dfrac{2\cdot \, 4\cdot 3\cdot 2\cdot 1}{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=\dfrac{2}{7\cdot 6\cdot 5}=\dfrac{1}{105}\\\\\\2)\ \ f(n)=\dfrac{(2n^2-3)!}{(n^2-1)!\, 9!}$

$f(3)=\dfrac{(2\cdot 9-3)!}{(9-1)!\, 9!}=\dfrac{15!}{8!\, 9!}=\dfrac{9!\cdot 10\cdot 11\cdot 12\cdot 13\cdot 14\cdot 15}{8!\cdot 9!}=\dfrac{10\cdot 11\cdot 12\cdot 13\cdot 14\cdot 15}{8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=\\\\\\=\dfrac{5\cdot 11\cdot 13}{8}=\dfrac{715}{8}$

$3)\ \ f(n)=\dfrac{(n^2-2n-3)!}{(3n-1)!(n+2)!}\\\\\\f(5)=\dfrac{12!}{14!\cdot 6!}=\dfrac{12!}{12!\cdot 13\cdot 14\cdot 7!}=\dfrac{1}{182\cdot 5040}=\dfrac{1}{917280}$

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