2. Find the sum of the following series. Round to the nearest hundredth if necessary. \[ 6+12+24+\ldots+1536 \] Sum of a finite geometric series: \[ S_{n}=\frac{a_{1}-a_{1} r^{n}}{1-r} \]

Step 1 ：Given the geometric series \(6+12+24+\ldots+1536\)

Step 2 ：We can see that the common ratio (r) is 2 (each term is twice the previous term) and the first term (a1) is 6.

Step 3 ：The number of terms (n) can be found by dividing the last term by the first term and taking the base-2 logarithm, then adding 1. So, \(n = \log_{2}(\frac{1536}{6}) + 1 = 9.0\)

Step 4 ：We can use the formula for the sum of a geometric series to find the sum: \(S_{n} = \frac{a_{1}(1 - r^{n})}{1 - r}\)

Step 5 ：Substitute the values into the formula: \(S_{n} = \frac{6(1 - 2^{9})}{1 - 2} = 3066.0\)

Step 6 ：Final Answer: The sum of the series is \(\boxed{3066}\)