Respuesta :
Answer:
The value of k is greater than or equal to 0, i.e. k≥7.
Step-by-step explanation:
The given equation is
[tex]k!+48=48((k+1)^m)[/tex]
The value of k must be a positive integer because k! is defined for k≥0, where k∈Z.
Subtract 48 from both the sides.
[tex]k!=48((k+1)^m)-48[/tex]
[tex]k!=48((k+1)^m-1)[/tex]
[tex]k!=48(k+1-1)(\frac{(k+1)^m-1}{(k+1)-1})[/tex]
Using [tex][\frac{r^m-1}{r-1}=r^{m-1}+r^{m-2}+...+1][/tex], we get
[tex]k!=48k((k+1)^{m-1}+(k+1)^{m-2}+...+1)[/tex]
Divide both sides by 48k.
[tex]\frac{k!}{48k}=(k+1)^{m-1}+(k+1)^{m-2}+...+1[/tex]
[tex]\frac{k(k-1)!}{48k}=(k+1)^{m-1}+(k+1)^{m-2}+...+1[/tex]
[tex]\frac{(k-1)!}{48}=(k+1)^{m-1}+(k+1)^{m-2}+...+1[/tex]
Note: The value of m can be 0 or 1.
The value of k is positive integer, so the right hand side of the above equation must be a positive integer.
Since RHS of the equation is positive integer, therefore (k-1)! is completely divisible by 48.
[tex]k-1\geq 6[/tex]
Add 1 on both sides.
[tex]k\geq 6+1[/tex]
[tex]k\geq 7[/tex]
Therefore the value of k is greater than or equal to 0.
