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# Help needed

Prove the following:

$$2^{k} > k^{3}$$ , $$\forall$$ $$k>9$$

Note by Swapnil Das
1 year, 3 months 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 $$2^{k+1}>(k+1)^3+(k-9)(0) = (k+1)^3 \forall \ k > 9$$. · 1 year, 3 months ago

Nihar really Impressive solution · 1 year, 3 months ago

Thanks! :) · 1 year, 3 months ago

@Nihar Mahajan

How is one supposed to think so beautifully? · 1 year, 3 months ago

Thanks! :) · 1 year, 3 months ago