Răspuns:
Ecuatia nu are solutii.
Explicație pas cu pas:
C₂ₙⁿ⁻¹ = 5·C₂ₙ₋₁ⁿ
Conditii ; n ∈ N }
2n ≥ n-1 => n ≥ -1 } => n ≥ 1 => n ∈ N*
2n-1 ≥ n => n ≥ 1 }
C₂ₙⁿ⁻¹ = 5·C₂ₙ₋₁ⁿ <=>
(2n)! (2n-1)!
___________ = 5·_____________ <=>
(n-1)!·(2n-n+1)! n!·(2n-1-n)!
(2n)! (2n-1)!
___________ = 5·_____________ <=>
(n-1)!·(n+1)! n!·(n-1)!
(2n)! · n! = 5·(n+1)!·(2n-1)! <=>
(2n)·n!·(2n-1)! = 5·(n+1)·n!·(2n-1)! I:[n!·(2n-1)!] =>
2n = 5(n+1) <=> 2n = 5n+5 => 3n = -5 =>
Ecuatia nu are solutii.
