Answer
$1^{k}+2^{k}+\ldots+n^{k}$ is $O\left(n^{k+1}\right)$ with $k=1$ and $C=1$.
Work Step by Step
We have given that $k$ is a positive integer
in order to show that $1^{k}+2^{k}+\ldots+n^{k}$ is $O\left(n^{k+1}\right)$
for convenience sake, we choose $k=1$ and $C=1$.
Whenever $x>k=1$, we then obtain:
$\begin{aligned}\left|1^{k}+2^{k}+\ldots+n^{k}\right| &=1^{k}+2^{k}+\ldots+n^{k} \\ & \leq n^{k}+n^{k}+\ldots+n^{k} \end{aligned}$
Note that the sum $1^{k}+2^{k}+\ldots+n^{k}$ contains $n$ terms and thus the sum $n^{k}+n^{k}+\ldots+n^{k}$ contains $n$ terms with each term $n^{k}$.
$=n$$ \cdot n^{k}$
$=n^{k+1}$
$=\left|n^{k+1}\right|$
By the Big- O definition, we then obtained that $1^{k}+2^{k}+\ldots+n^{k}$ is $O\left(n^{k+1}\right)$ with $k=1$ and $C=1$.
