Q13. Prove algebraically that the difference between the squares of any two consecutive odd numbers is always a multiple of 8

## Step1:By definition, any odd number on a number line can be expressed as or , where 'n' is any integer and each term accounts for one possible sequence of odd numbers (that is, {1,3,5,7...} or {2, 4, 6, 8...}). For this particular task, we need to focus on consecutive odd numbers. Let's assign typical consecutive odd numbers such as 'n' & 'n+2'.In other words, we can algebraically represent two consecutive odd numbers like: (old version) and (next version). Notice the interval is due to twice the value in between any two consecutive odd numbers.## Step2:To prove our initial claim, we need to find the difference of the squares of these two consecutive odd integers...\((2n + 1)^2 - (2n - 1)^2 \). Instead of just multiplying this out as is, notice that this formula falls under the Difference of Squares pattern \((a–b)(a + b)\). Here, and .## Step3:Next, we continue with factoring the difference of squares expression. Notice that we are not interested in \(2n + 1 - (2n - 1) = 2\); that's another possibility. But rather factoring by adding both 'a' and 'b' in this context... . Meaning (\((2n + 1) + (2n - 1)\) equals ### \((2n + 1)^2 - (2n - 1)^2 = 4n *(2n + 1 + 2n - 1) = 4n *(4n) = 16n^2\)The final expression shows that the difference of squares of any two consecutive odd integers is equal to , which is a multiple of 8. Thus our initial claim is proved algebraically.