Prove that 2 divides k(k+1) for all k?

K is in the universe of natural numbers.

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  • Anonymous
    10 years ago
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    Since k is a natural number, then it's either even or odd. If k is even, then we can let k = 2m for some other natural number "m". Then k(k+1) = 2 * m(k+1), so 2 clearly divides it. If k is odd, then k+1 is even, and we can likewise let k+1 = 2m to get k(2m) which is divisible by 2. So in both cases, k(k+1) is even and thus it should always be divisible by 2 for any natural k.

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