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Prove that $16 + 3^n=O(4^{n})$. I have tried to do this problem but cannot find a constant $c$ that I am supposed to find.

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Notice that For $n\ge 3$, $3^n+16< 4^n$. You can show it by induction if you want

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$$16+3^n\le17(4^n)\text{ for all } n\in\mathbb{R^+}$$ So we can use $c=17$ if we want the Big-O notation to be valid for all positive $n$.

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Argue that $$ \frac{16+3^n}{4^n}\to 0 $$ as $n\to \infty$ and hence that $16+3^n<4^n$ for sufficiently large $n$ by the definition of the limit.

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