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Please help me with this problem based on Induction.

Prove the following:

\(2^{k} > k^{3}\) , \(\forall\) \(k>9\)

Note by Swapnil Das
11 months, 3 weeks ago

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Induction base: \(2^{10}=1024>1000=10^3\) . Thus the inequality holds for \(k=10\).

Induction step: Lets assume that \(2^k > k^3 \forall k>9\).

Thus , \(2^{k+1}=2\times 2^k > 2k^3\).

Note that \(2k^3 = (k+1)^3+k^3-3k^2-3k-1\). Hence ,

\[\begin{align} 2^{k+1} &> (k+1)^3+k^3-3k^2-3k-1 \\ &> (k+1)^3+k^3-3k^2-3k-459 \\ &= (k+1)^3+(k-9)(k^2+6k+51) \end{align}\]

Note that \(k^2+6k+51 > 0\) since its discriminant is negative.

Thus finally we have \(2^{k+1}>(k+1)^3+(k-9)(0) = (k+1)^3 \forall \ k > 9\). Nihar Mahajan · 11 months, 3 weeks ago

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@Nihar Mahajan Nihar really Impressive solution Atul Shivam · 11 months, 3 weeks ago

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@Atul Shivam Thanks! :) Nihar Mahajan · 11 months, 3 weeks ago

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@Nihar Mahajan @Nihar Mahajan

How is one supposed to think so beautifully? Swapnil Das · 11 months, 3 weeks ago

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@Swapnil Das Thanks! :) Nihar Mahajan · 11 months, 3 weeks ago

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Yup, definitely check out Induction. This is covered under the inequalities section. Calvin Lin Staff · 11 months, 3 weeks ago

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