#### Answer

The sum is $3280$.

#### Work Step by Step

The given geometric sequence has the following terms:
$1, 3, 9, ..., 2187$
The sequence has:
$a=1
\\r=\dfrac{3}{1}=3$
RECALL:
The partial sum $S_n$ (sum of the first $n$ terms) of a geometric sequence is given by the formula:
$S_n=a\left(\dfrac{1-r^n}{1-r}\right), r\ne 1$
where
$a$ = first term
$r$ = common ratio
As of now, the values of $a$ and $r$ are known.
However, the term number of $2187$ is unknown.
Thus, we cannot solve for the partial sum yet.
We first have to know which term of the sequence is $2187$.
The $n^{th}$ term of a geometric sequence is given by the formula $a_n = a \cdot r^{n-1}$ where $a$ = first term and $r$ = common ratio.
Thus, the $n^{th}$ term of the sequence is:
$a_n = 1 \cdot 3^{n-1}$
To know which term of the sequence is 2187, substitute 2187 to $a_n$ in the formula above to obtain:
$a_n = 1 \cdot 3^{n-1}
\\2187 = 1 \cdot 3^{n-1}
\\2187 = 3^{n-1}$
Note that $2187 = 3^7$. This means that the equation above is equivalent to:
$3^7 = 3^{n-1}$
Use the rule $a^m = a^n \longrightarrow m=n$ to obtain:
$7=n-1
\\7+1=n-1+1
\\8=n$
Thus, $a_8=2187$.
This means that the partial sum involved is the sum of the first eight terms.
Use the formula for the partial sum to obtain:
$\require{cancel}
S_8 =1\left(\dfrac{1-3^8}{1-3}\right)
\\S_8=\left(\dfrac{1-6561}{-2}\right)
\\S_8=\left(\dfrac{-6560}{-2}\right)
\\S_8=3280$