Respuesta :
n=an even integer.
the next even number is: (n+2)
the next even number to (n+2) is: (n+2)+2=n+4
the next even number to (n+4) is : (n+4)+2=n+6
Therefore, the sumf of thant number and the next even numbers afeter it will be:
n + (n+2)+(n+4)+(n+6)
if we simplifi this expression fully, we have:
n + (n+2)+(n+4)+(n+6)=4n+12
Answer: Let "n" an even integer, the expression for the sum of tahta number and the next three even numbers after it , will be:
4n+12
the next even number is: (n+2)
the next even number to (n+2) is: (n+2)+2=n+4
the next even number to (n+4) is : (n+4)+2=n+6
Therefore, the sumf of thant number and the next even numbers afeter it will be:
n + (n+2)+(n+4)+(n+6)
if we simplifi this expression fully, we have:
n + (n+2)+(n+4)+(n+6)=4n+12
Answer: Let "n" an even integer, the expression for the sum of tahta number and the next three even numbers after it , will be:
4n+12
Answer:
The required expression is 4n+12.
Step-by-step explanation:
Consider the provided information.
We need to write an expression for the sum of that number and the next three even numbers after it.
Let n represent an even integer.
The next three even numbers after it is:
First even number: n
The second even number: n+2
Third even number: n+4
Fourth even number: n+6
The sum of all even number is:
n+n+2+n+4+n+6
Add the like terms:
4n+6+6
4n+12
Hence, the required expression is 4n+12.
