To prove that the difference between the squares of consecutive even numbers is always a multiple of 4, let's proceed step-by-step with the algebraic expressions.
### Step 1: Define Consecutive Even Numbers
Let [tex]\( n \)[/tex] be any integer. We can define two consecutive even numbers using [tex]\( n \)[/tex] as follows:
- The first even number: [tex]\( 2n \)[/tex]
- The next consecutive even number: [tex]\( 2n + 2 \)[/tex]
### Step 2: Express the Squares of These Numbers
Now, we'll compute the square of each of these even numbers:
- Square of the first even number: [tex]\( (2n)^2 = 4n^2 \)[/tex]
- Square of the next consecutive even number: [tex]\( (2n + 2)^2 = (2n + 2)(2n + 2) = 4n^2 + 8n + 4 \)[/tex]
### Step 3: Find the Difference Between the Squares
Next, we need to find the difference between the square of the second even number and the square of the first even number:
[tex]\[ \text{Difference} = (2n + 2)^2 - (2n)^2 \][/tex]
Substitute the expressions we computed:
[tex]\[ \text{Difference} = (4n^2 + 8n + 4) - 4n^2 \][/tex]
### Step 4: Simplify the Difference
Simplify the expression by combining like terms:
[tex]\[ \text{Difference} = 4n^2 + 8n + 4 - 4n^2 \][/tex]
[tex]\[ \text{Difference} = 8n + 4 \][/tex]
### Step 5: Prove That the Difference is a Multiple of 4
The next step is to show that the simplified difference, [tex]\( 8n + 4 \)[/tex], is always a multiple of 4. We can factor out 4 from the expression:
[tex]\[ 8n + 4 = 4(2n + 1) \][/tex]
Since [tex]\( 2n + 1 \)[/tex] is an integer (because [tex]\( n \)[/tex] is an integer), it is evident that [tex]\( 8n + 4 \)[/tex] is always a multiple of 4.
### Conclusion
Therefore, we have shown algebraically that the difference between the squares of two consecutive even numbers is always a multiple of 4.
