Find the sum of the following series.Round to the nearest hundredth if necessary. 2+12+72+ldots +120932352 Sum of a finite geometric series: S_(n)=(a_(1)-a_(1)r^n)/(1-r)

To find the sum of the given series, we need to identify the first term, common ratio, and the number of terms in the series.<br /><br />The first term \(a_1\) is 2.<br /><br />The common ratio \(r\) can be found by dividing the second term by the first term:<br />\[ r = \frac{12}{2} = 6 \]<br /><br />Next, we need to determine the number of terms \(n\) in the series. We know the last term is 120932352. The \(n\)-th term of a geometric series is given by:<br />\[ a_n = a_1 \cdot r^{n-1} \]<br /><br />So, we set up the equation:<br />\[ 120932352 = 2 \cdot 6^{n-1} \]<br /><br />Solving for \(n\):<br />\[ 6^{n-1} = \frac{120932352}{2} \]<br />\[ 6^{n-1} = 60466176 \]<br /><br />Taking the logarithm of both sides:<br />\[ n-1 = \log_6{60466176} \]<br /><br />Using a calculator to find the logarithm:<br />\[ n-1 \approx \log_6{60466176} \approx 8.58 \]<br /><br />Since \(n\) must be an integer, we round to the nearest whole number:<br />\[ n \approx 9 \]<br /><br />Now, we can use the formula for the sum of a finite geometric series:<br />\[ S_n = \frac{a_1 - a_1 r^n}{1 - r} \]<br /><br />Substituting the values:<br />\[ S_9 = \frac{2 - 2 \cdot 6^9}{1 - 6} \]<br /><br />Calculating the numerator:<br />\[ 2 - 2 \cdot 6^9 = 2 - 2 \cdot 10077696 = 2 - 20155392 = -20155390 \]<br /><br />Now, the denominator:<br />\[ 1 - 6 = -5 \]<br /><br />So, the sum is:<br />\[ S_9 = \frac{-20155390}{-5} = 4031078 \]<br /><br />Therefore, the sum of the series is:<br />\[ \boxed{4031078} \]