Aryan1078 Aryan1078

01-01-2018
Mathematics

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Prove that n^2 > 2n + 1 for all natural numbers n >= 3

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01-01-2018

[tex] {n}^{2} - 2n - 1 > 0[/tex]
[tex]n = 3 > > 9 - 6 - 1 = 2 > 0 [/tex]
suppose the following is true

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