I need to prove that: $2^{2n}-n^2+3^n = \Omega (2^{2n})$
I started and got to this: $2^{2n}-n^2+3^n \geq 2^{2n}\cdot 3 \geq 2^{2n}\cdot 2 = 2^{2n+1}$ for every $n > n_{0} = 1$
How should I continue from here?
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If $n=4$, $2^{2\cdot4}-4^2+3^4 = 321$ and $2^{2\cdot4}\cdot 3=768$, so the first part of your inequality ($2^{2n}-n^2+3^n \geq 2^{2n}\cdot 3$) is not correct. – Steve Kass Mar 29 '16 at 19:37
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1$2^{2n} - n^2 + 3^n > 2^{2n}$ for non-negative $n$ – Klint Qinami Mar 29 '16 at 19:43
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Assume $n \in \mathbb{R}$ and $n > 0$.
$$2^{2n} - n^2 + 3^n > 2^{2n}$$ $$3^n > n^2$$
This should be relatively easy to show right?
Klint Qinami
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