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anonymous
 one year ago
Show that n^3 > (n+1)^2 for all natural numbers n>=3
anonymous
 one year ago
Show that n^3 > (n+1)^2 for all natural numbers n>=3

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5\(n\ge 3\) multiply both sides by \(n^2\) and get \(n^3\ge 3n^2\\=n^2+n^2+n^2 \\=n^2 + n\cdot n+n^2 \\\gt n^2 + 2\cdot n + 1 \\= (n+1)^2\)

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2@ganeshie8 : brilliant! Or if we prefer the long way, using mathematical induction, Statement: \(n^3>(n+1)^2\) for n>=3. Base case: for n=3, we have \(n^3(n+1)^2=3^34^2=11 > 0\) so statement holds for n=3. Inductive Hypothesis: assuming statement holds for case n=k. i.e. \(k^3(k+1)^2 = S_k >0\) Inductive step: for case n=k+1, we have \((k+1)^3(k+1+1)^2 =k^3+3k^2+3k+1(k^2+4k+4)\) \(=[k^3(k^2+2k+1)] + 3k^2+k2)\) \(=S_k +(3k2)(k+1)\) On the last line, \(S_k >0 \) by hypothesis (3k2) > 0 \(\forall k\ge3\) (k+1) >0 \(\forall k\ge3\) Therefore \(=S_k +(3k2)(k+1) > 0\) for case n=k+1 Conclusion: By mathematical induction, we have established that statement \(n^3>(n+1)^2\) is true for all integers n>=3
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