Solve using the definition of Big-$\mathcal{O}$ By using the definition of Big-$\mathcal{O}$, show that

\begin{align} 2^{(n+2)} + 3^{(n+1)} & \text{ is } \mathcal{O}(3^n)\\ \sqrt{10n^2+7n+3} & \text{ is } \mathcal{O}(n)\end{align}

I'm not sure where to start with this, especially with the last one. Can someone help please?
 A: $$\overbrace{2^{n+2} + 3^{n+1}  < 3^{n+2} + 3^{n+1}}^{(\because \,\, 2 < 3)} = 12 \cdot 3^{n}$$
$$\underbrace{\sqrt{10n^2 + 7n+3} \leq \sqrt{10n^2 + 7n^2+3n^2}}_{\because \,\, n \leq n^2 \,\, \& \,\, 1 < n^2, \,\,\, \forall n \in \mathbb{Z}^+} = \sqrt{20n^2} = \sqrt{20} \cdot n $$
A: More generally,
if the $a_k$ are $m$ real numbers
and $A = \max(a_k)$,
then,
for any positive real $r$,
$\sum a_k^r = O(A^r) 
$
If, in addition,
the $b_k$ are $m$ positive real numbers
and $c$ is a positive real,
$(\sum b_k r^{a_k})^c = O(r^{Ac})
$.
Both of these are with the $O$ being as $r \to \infty$
Both of these can be proved by
dividing by the claimed bound
and showing that the result is
$O(1)$.
A: By the definition of the Big-$O$ function:
$$ \left| 2^{n+2} + 3^{n+1} \right| \le A \cdot \left| 3^n \right| $$
$$ \left| 4 \cdot 2^n + 3 \cdot 3^n  \right| \le A \left| 3^n \right| $$
By the triangular inequality:
$$ \left| 4 \cdot 2^n + 3 \cdot 3^n \right| \le \left| 4 \cdot 2^n \right| + \left| 3 \cdot 3^n \right| $$
We can show that for all $n \ge 0$ (since $3>2$):
$$ \left| 4 \cdot 2^n \right| + \left| 3 \cdot 3^n \right| \le \left| 4 \cdot 3^n \right| + \left| 3 \cdot 3^n \right| $$
So we have:
$$ 7 \cdot \left| 3^n \right| \le A \cdot \left| 3^n \right| $$
$$ 7 \le A $$

For $n$ sufficiently large (viz. $n\ge\frac{7 + \sqrt{61}}{2}$), we have:
$$ \sqrt{10n^2 + 7n + 3} \le \sqrt{11n^2} \le A \cdot n$$
$$ \sqrt{11} \le A $$
You can show that $11 n^2 > 10n^2 + 7n + 3$ by solving the quadratic appropriately to get that it is true for $n$ not between $\frac{7 \pm \sqrt{61}}{2}$.
