Question

preciso resolver o seguinte problema: 5.(n-3)!+5.(n-1)!=31.(n-2)!

Answer 1

$5(n - 3)! + 5(n - 1)! = 31(n - 2)! \\ 5(n - 3)! + 5(n - 1)(n - 2)(n - 3)! = 31(n - 2)(n - 3)! \;\;\;\; \div[(n - 3)! \\ 5 + 5(n - 1)(n - 2) = 31(n - 2) \\ 5[1 + (n - 1)(n - 2)] = 31(n - 2) \\ 5(1 + n^2 - 3n + 2) = 31n - 62 \\ 5n^2 - 15n - 31n + 15 + 62 = 0 \\ 5n^2 - 46n + 77 = 0 \\ \Delta = 2116 - 1540 \\ \Delta = 576 \\\\ x = \frac{46 \pm \sqrt{576}}{10} \begin{cases} x' = \frac{46 + 24}{10} \Rightarrow \boxed{\boxed{x' = 7}} \\ x'' = \frac{46 - 24}{10} \Rightarrow x'' = 2,2 \end{cases}$

Answer 2

Temos que:

 

$5\cdot(\text{n}-3)!+5\cdot(\text{n}-1)!=31\cdot(\text{n}-2)!$

 

Observe que:

 

$(\text{n}-2)!=(\text{n}-2)\cdot(\text{n}-3)!$

 

$(\text{n}-1)!=(\text{n}-1)\cdot(\text{n}-2)\cdot(\text{n}-3)!$

 

Desta maneira, podemos escrever:

 

$5\cdot(\text{n}-3)!+5\cdot(\text{n}-1)\cdot(\text{n}-2)\cdot(\text{n}-3)!=31\cdot(\text{n}-2)\cdot(\text{n}-3)!$

 

Dividindo ambos os membros por $(\text{n}-3)!$:

 

$5+5\cdot(\text{n}-1)\cdot(\text{n}-2)=31\cdot(\text{n}-2)$

 

$5+5\text{n}^2-15\text{n}+10=31\text{n}-62$

 

$5\text{n}^2-46\text{n}+77=0$

 

$\text{n}=\dfrac{-(-46)\pm\sqrt{(-46)^2-4\cdot5\cdot77}}{2\cdot5}=\dfrac{46\pm24}{10}$

 

Logo, as raízes são:

 

$\text{n}'=\dfrac{46+24}{10}=7$

 

$\text{n}"=\dfrac{46-24}{10}=\dfrac{11}{5}$

 

Como $\text{n}\in\mathbb{Z}^+$, segue que $\text{n}=7$.