Question 2

We will represent the consecutive odd numbers algebraically and simplify the expression.

Step 1 — Representing the numbers

Let's pick an integer $n$. We can write any odd number as $2n - 1$. The next consecutive odd number is $2n + 1$. We will use these two expressions.

Step 2 — Sum of their squares

First, we square the first odd number. $(2n - 1)^2$ $= (2n)^2 - 2(2n)(1) + 1^2$ $= 4n^2 - 4n + 1$

Next, we square the second odd number. $(2n + 1)^2$ $= (2n)^2 + 2(2n)(1) + 1^2$ $= 4n^2 + 4n + 1$

Now, we add these two squared results together. $(4n^2 - 4n + 1) + (4n^2 + 4n + 1)$ $= 4n^2 + 4n^2 - 4n + 4n + 1 + 1$ $= 8n^2 + 2$

$\boxed{8n^2 + 2}$

Step 3 — Add 6 to the result

We take the sum of squares and add 6 to it. $(8n^2 + 2) + 6$ $= 8n^2 + 8$

$\boxed{8n^2 + 8}$

Step 4 — Proving divisibility by 8

Let's look at the new number we found. We can factor out a common term. $8n^2 + 8$ $= 8(n^2 + 1)$

Since $n$ is an integer, $n^2$ is an integer. So, $n^2 + 1$ is also an integer. This means the entire expression is 8 multiplied by an integer. Any number that can be written as 8 times an integer is divisible by 8.

Answer

(i) The algebraic representation of two consecutive odd numbers is $2n-1$ and $2n+1$. (ii) The sum of their squares plus 6 is $8(n^2 + 1)$. (iii) Since the expression is $8$ multiplied by an integer, it is always divisible by $8$.