Respuesta :

ghanami

Answer:

Step-by-step explanation:

hello :

let four consecutive whole numbers :   n  ,  n+1 , n+2  , n+3

the sum is :  n  +  n+1 + n+2 + n+3 =4n+6

you can write 4n+6 to the form : 2(2n+3)    means   2k    ( k=2n+3)

so even number  

vijayhalder031

Since the sum of any four consecutive integers can be written in the form  2(2n+3) where  n  is the first integer in the series, we conclude that this sum is always an even number

Whole numbers

  • A whole number is simply any positive number that does not include a fractional or decimal part
  • In mathematics, whole numbers are the basic counting numbers 0, 1, 2, 3, 4, 5, 6, …, and so on.
  • Whole numbers is a collection of positive integers and zero or we can say that whole numbers are the set of non-negative integers.

How to solve this problem?

The steps are as follow:

  • We suppose that the first number in our sequence is the integer n.
  • Then, our sequence of consecutive numbers can be written as: n, n+1, n+2, n+3
  • We then sum these together: S=(n)+(n+1)+(n+2)+(n+3)
  • Collect the terms together:

S=(n+n+n+n)+(1+2+3)

S=4×n+6

  • We can now pull out a factor of two: S=2×(2n+3)
  • The final thing to note is that the definition of an even number is a number p such that: p=2×m,   m∈Z

Therefore, since the sum of any four consecutive integers can be written in the form  2(2n+3) where  n  is the first integer in the series, we conclude that this sum is always an even number.

Learn more about Whole numbers here:

https://brainly.com/question/251701

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