Prove using algebra that the difference between the squares of consecutive odd numbers is always a multiple of 8.

Answer:

See  below

Step-by-step explanation:

Lets label the consecutive odd numbers as 2n+1 and 2n+3 where n is an integer.

We need to prove that

((2n + 3)^2 - (2n + 1)^2 is a multiple of 8.

Using the difference of 2 squares on the left side:-

(2n + 3 + 2n + 1)(2n + 3 -  (2n + 1)

=  (4n + 4)(2)

= 8n + 8  which is a multiple of 8  

Final answer:

The difference between squares of consecutive odd numbers can be expressed as 4n+4 which is a multiple of 8.

Explanation:

Let's take two consecutive odd numbers. We can express an odd number as 2n+1 where n is any integer. So, the consecutive odd number will be 2n+3 (adding 2 to the previous one).

Now, the squares of these consecutive numbers are (2n+1)² and (2n+3)² respectively.

The difference between these squares is (2n+3)² - (2n+1)². Applying the formula a² - b² = (a+b)(a-b), we get 4n+4 as the answer which is a multiple of 8 as 8 divides it without a remainder.

So the difference between the squares of consecutive odd numbers is a multiple of 8.

Learn more about Consecutive Odd Numbers here:

brainly.com/question/35983904

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