Question Find the sum of the first 6 terms of the following sequence. Round to the nearest hundredth if necessary. $$ 8, \quad 4, \quad 2, \ldots $$ Sum of a finite geometric series: $$ S_{n}=\frac{a_{1}-a_{1} r^{n}}{1-r} $$

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Let the first term be $$ a_{1}=8 $$ and the common ratio be
$$
r=\frac{4}{8}=\frac{1}{2}.
$$
The sum of the first 6 terms is given by:
$$
S_{6}=\frac{a_{1}-a_{1}r^{6}}{1-r}.
$$
Substitute $$ a_{1}=8 $$ and $$ r=\frac{1}{2} $$:
$$
S_{6}=\frac{8-8\left(\frac{1}{2}\right)^{6}}{1-\frac{1}{2}}.
$$
Calculate $$\left(\frac{1}{2}\right)^{6}$$:
$$
\left(\frac{1}{2}\right)^{6}=\frac{1}{64}.
$$
So,
$$
S_{6}=\frac{8-8\left(\frac{1}{64}\right)}{\frac{1}{2}}=\frac{8-\frac{8}{64}}{\frac{1}{2}}=\frac{8-\frac{1}{8}}{\frac{1}{2}}.
$$
Convert $$8$$ to a fraction with denominator $$8$$:
$$
8=\frac{64}{8}.
$$
Then,
$$
\frac{64}{8}-\frac{1}{8}=\frac{63}{8}.
$$
Now, divide by $$\frac{1}{2}$$:
$$
S_{6}=\frac{\frac{63}{8}}{\frac{1}{2}}=\frac{63}{8}\times2=\frac{63}{4}.
$$
Finally, compute the fraction:
$$
\frac{63}{4}=15.75.
$$

The sum of the first 6 terms is $$15.75$$.
The sum of the first 6 terms is 15.75.

Question Find the sum of the first 6 terms of the following sequence. Round to the nearest hundredth if necessary. \[ 8, \quad 4, \quad 2, \ldots \] Sum of a finite geometric series: \[ S_{n}=\frac{a_{1}-a_{1} r^{n}}{1-r} \]

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