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5 November, 18:39

Prove that the difference between squares of consecutive even numbers is always a multiple of 4.

Note: Let n stand for any integer in your working.

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  1. 5 November, 19:59
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    Proved: the difference between squares of consecutive even numbers is always a multiple of 4.

    Step-by-step explanation:

    Let's consider two consecutive even numbers.

    If n = a positive integer

    Then the 1st even number would be = 2n

    Let the next consecutive even number = 2 (n+1)

    The consecutive even numbers would be:

    2n, 2 (n+1)

    The square of each consecutive even number would be:

    (2n) ² = 2n * 2n = 4n²

    [2 (n+1) ]² = (2n+2) ² = (2n+2) (2n+2)

    Expanding the brackets

    = 4n² + 4n + 4n+4

    [2 (n+1) ]² = 4n²+8n+4

    To determine if the difference between squares of consecutive even numbers is always a multiple of 4, we would assign values for n and find the difference of the squares of each consecutive even numbers.

    Let n = 1, 2, 3

    For n = 1

    4n² = 4 (1) ² = 4*1 = 4

    4n²+8n+4 = 4 (1) ² + 8 (1) + 4

    = 4+8+4 = 16

    Difference between the squares of consecutive even numbers = 16-4 = 12

    For n = 2

    4n² = 4 (2) ² = 4*4 = 16

    4n²+8n+4 = 4 (2) ² + 8 (2) + 4

    = 16+16+4 = 36

    Difference between the squares of consecutive even numbers = 36-16 = 20

    For n = 3

    4n² = 4 (3) ² = 4*9 = 36

    4n²+8n+4 = 4 (3) ² + 8 (3) + 4

    = 36+24+4 = 64

    Difference between the squares of consecutive even numbers = 64-36 = 28

    12 = 4*3

    20 = 4*5

    28 = 4*7

    From the results above, we can see that 12, 20, and 28 are multiples of 4.

    Therefore, the difference between squares of consecutive even numbers is always a multiple of 4.
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