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31 October, 18:54

True or false. If a is any odd integer, then a^2 + a is even. Explain this.

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  1. 31 October, 19:22
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    True.

    Step-by-step explanation:

    We can represent an odd number by 2n + 1 where n = 0, 1, 2, 3, 5 etc.

    Substituting:

    a^2 + a = (2n + 1) ^2 + 2n + 1

    = 4n^2 + 4n + 1 + 2n + 1

    = 4n^2 + 6n + 2

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

    which is even because any integer multiplied by an even number is even.

    This is also true if we use a negative odd integer:

    We have 4n^2 + 4n + 1 - 1 - 2n

    = 4n^2 + 2n

    = 2 (2n^2 + n (.
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