Question

a,b,c,d,e va rog ofer coroana

Answer 1

• Factorialul lui n este notat cu n! și este egal cu produsul numerelor naturale nenule mai mici sau egale cu n

       $\boxed{1 \cdot 2 \cdot 3 \cdot ... \cdot n = n!}$

a)

$(n + 3)!(n + 4 - n) = 96 \Rightarrow 4(n+3)! = 96 \ \ \Big|:4$

$\Rightarrow (n+3)! = 24 = 4! \Rightarrow n+3=4 \Rightarrow n = 1$

$\implies S = \Big\{1\Big\}$

b)

$3(n+1)! = (n+2)! \Rightarrow 3(n+1)! = (n+2)(n+1)! \ \ \Big|:(n+1)!$

$\Rightarrow n+2=3 \Rightarrow n = 1$

$\implies S = \Big\{1\Big\}$

c)

$(n+4)!(n+2) = 48(n+3)! \Rightarrow (n+3)!(n+4)(n+2) = 48(n+3)! \ \ \Big|:(n+3)!$

$\Rightarrow (n+4)(n+2)=48 \Rightarrow n^{2} + 6n - 40 = 0$

$\Rightarrow (n+10)(n-4) = 0 \Rightarrow n_{1} = -10 \not \in \mathbb{N}; \ n_{2} = 4$

$\implies S = \Big\{4\Big\}$

d)

$3(n+2)!+5(n+1)!=51n! \Rightarrow 3(n+2)(n+1)n!+5(n+1)n!=51n! \ \ \Big|:n!$

$\Rightarrow 3(n+2)(n+1)+5(n+1)=51 \Rightarrow 3n^{2} + 14n - 40 = 0$

$\Rightarrow (3n + 20)(n - 2) = 0 \Rightarrow n_{1} = \dfrac{20}{3} \not \in \mathbb{N}; \ n_{2}= 2$

$\implies S = \Big\{2\Big\}$

e)

$(n+3)!+18(n+1)! = 10(n+2)!$

$\Rightarrow (n+3)(n+2)(n+1)!+18(n+1)! = 10(n+2)(n+1)! \ \ \Big|:(n+1)!$

$\Rightarrow (n+3)(n+2)+18 = 10(n+2) \Rightarrow n^{2} - 5n + 4 = 0$

$\Rightarrow (n-1)(n-4)=0 \Rightarrow n_{1} = 1; \ n_{2} = 4$

$\implies S = \Big\{1;4\Big\}$