## College Algebra 7th Edition

The sum is $3280$.
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: $\require{cancel} S_8 =1\left(\dfrac{1-3^8}{1-3}\right) \\S_8=\left(\dfrac{1-6561}{-2}\right) \\S_8=\left(\dfrac{-6560}{-2}\right) \\S_8=3280$