Question

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)-?

Answer 1

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

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

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

Answer 2

Let's calculate the values of the given functions:

1. f(n) = 2n! / (2n-1)! To calculate f(4), we substitute n = 4 into the formula: f(4) = 2(4)! / (2(4)-1)! = 2(4 x 3 x 2 x 1) / (7 x 6 x 5 x 4 x 3 x 2 x 1) = 24 / (7 x 6 x 5 x 4 x 3 x 2 x 1) = 24 / 5040 = 1/210

   Therefore, f(4) = 1/210.

2. f(n) = (2n^2 - 3)! / (n^2 - 1)! * 9! To calculate f(3), we substitute n = 3 into the formula: f(3) = (2(3)^2 - 3)! / (3^2 - 1)! * 9! = (2(9) - 3)! / (9 - 1)! * 9! = (18 - 3)! / 8! * 9! = 15! / 8! * 9! = (15 x 14 x 13 x 12 x 11 x 10 x 9 x 8!) / 8! * 9! = (15 x 14 x 13 x 12 x 11 x 10 x 9) = 14414400

   Therefore, f(3) = 14414400.

3. f(n) = (n^2 - 2n - 3)! / (3n - 1)! * (n + 2)! To calculate f(5), we substitute n = 5 into the formula: f(5) = (5^2 - 2(5) - 3)! / (3(5) - 1)! * (5 + 2)! = (25 - 10 - 3)! / (15 - 1)! * (7)! = (12)! / (14)! * (7)! = (12 x 11 x 10 x 9 x 8 x 7!) / (14 x 13 x 12 x 11 x 10 x 9 x 8)! = 1 / (14 x 13) = 1/182

   Therefore, f(5) = 1/182.