Answer
$\sum_{k=1}^{33}(3k)$
Work Step by Step
Writing the given sum using sigma notation means to determine the general term of the sum and the limits for the summation index:
$$3+6+9+\dots+99=\sum_{k=m}^n a_k.$$
We look for the pattern in the terms of the sum:
$$\begin{align*}
3&=3\cdot 1\\
6&=3\cdot 2\\
9&=3\cdot 3\\
12&=3\cdot 4\\
&\dots\\
99&=3\cdot 33.
\end{align*}$$
So the term on the $k$th position is written as
$$a_k=3k.$$
The summation index $k$ goes from $1$ to $33$.
We conclude that the sum can be written using sigma notation like this:
$$3+6+9+12+\dots+99=\sum_{k=1}^{33}(3k).$$
