Question

va rog cu rezolvare ​

Answer 1

Explicație pas cu pas:

1) n ≥ 3, n ∈ ℕ

$\dfrac{6(2n)!}{(2n - 1)!} = \dfrac{n!}{(n - 3)!}$

$\dfrac{6(2n - 1)!2n}{(2n - 1)!} = \dfrac{(n - 3)!(n - 2)(n - 1)n}{(n - 3)!}$

$12n = (n - 2)(n - 1)n$

$n({n}^{2} - 3n - 10) = 0$

$n(n + 2)(n - 5) = 0$

$n = 0 \iff n \leqslant 3$

$n + 2 = 0 \iff n = - 2 \iff n \not \in \mathbb{N}$

$n - 5 = 0 \implies \bf n = 5$

2) n ≥ 1, n ∈ ℕ

$\dfrac{(3n)!}{(3n - 2)!} = \dfrac{5(n + 1)!}{(n - 1)!}$

$\dfrac{(3n - 2)!(3n - 1)3n}{(3n - 2)!} = \dfrac{5(n - 1)!n(n + 1)}{(n - 1)!}$

$(3n - 1)3n = 5n(n + 1)$

$n(9n - 3 - 5n - 5) = 0$

$n(4n - 8) = 0 \iff 4n(n - 2) = 0$

$n = 0 \leqslant 1$

$n - 2 = 0 \implies \bf n = 2$