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Prove by induction that for all n?N, n2 + n is an even integer (n = 1)

aniket anand , 11 Years ago
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Vikas TU

Last Activity: 5 Years ago

n^2 = (2k+1)^2 = 4k^2 +4k+1 = 2(2k^2 + 2k) + 1 which is also odd. Therefore if n is odd, then n^2 is odd. Thus, it follows that if n^2 is even, then n is even. If you assume that n must be an integer, then yes: if n^2 is even then n must be even.
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