Question 12 of 26intermediate🔧 ApplyLong Answer3 marks

Take two consecutive odd numbers. Find the sum of their squares, and then add 6 to the result. Prove that the new number is always divisible by 8.

Correct Answer

Let the two consecutive odd numbers be 2n+12n + 1 and 2n+32n + 3, where nn is an integer. The sum of their squares is:

(2n+1)2+(2n+3)2(2n + 1)^2 + (2n + 3)^2

According to the problem, we add 6 to this sum. The resulting expression is:

(2n+1)2+(2n+3)2+6(2n + 1)^2 + (2n + 3)^2 + 6

Expanding the squares, we get:

(4n2+4n+1)+(4n2+12n+9)+6(4n^2 + 4n + 1) + (4n^2 + 12n + 9) + 6

Simplifying the terms, we have:

8n2+16n+168n^2 + 16n + 16

Factoring out 8, this becomes 8(n2+2n+2)8(n^2 + 2n + 2). Since nn is an integer, (n2+2n+2)(n^2 + 2n + 2) is also an integer. Thus, the new number is always divisible by 8.

Exercise: EXERCISE A1.3 | Q: 2 | (Page 11)
For More Understanding

Explanation

The solution follows the algebraic proof methodology demonstrated in Example 11 of the provided context, where numbers are expressed in general algebraic forms (like 6k+16k+1). By representing consecutive odd numbers as 2n+12n+1 and 2n+32n+3, the proof proceeds through a logical sequence of steps—expansion, simplification, and factorization—to establish the divisibility property, adhering to the remark that each step must follow from previous steps.

Solution Steps

  1. Step 1: Let the consecutive odd numbers be 2n+12n + 1 and 2n+32n + 3.

  2. Step 2: Form the expression (2n+1)2+(2n+3)2+6(2n + 1)^2 + (2n + 3)^2 + 6.

  3. Step 3: Expand to get 4n2+4n+1+4n2+12n+9+64n^2 + 4n + 1 + 4n^2 + 12n + 9 + 6.

  4. Step 4: Simplify to 8n2+16n+168n^2 + 16n + 16.

  5. Step 5: Factor as 8(n2+2n+2)8(n^2 + 2n + 2) to prove divisibility by 8.

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