To calculate the sum of the geometric series \( \sum_{n=1}^{20} 12^{n-1} \), we first identify the parameters of the geometric series.
This series can be expressed in the form:
\[ S = a + ar + ar^2 + ar^3 + \ldots + ar^{n-1} \]
where:
- \( a = 12^0 = 1 \) (the first term),
- \( r = 12 \) (the common ratio),
- \( n = 20 \) (the number of terms).
The formula for the sum of the first \( n \) terms of a geometric series is given by:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
Plugging in our values:
- \( a = 1 \)
- \( r = 12 \)
- \( n = 20 \)
The formula becomes:
\[ S_{20} = 1 \cdot \frac{12^{20} - 1}{12 - 1} = \frac{12^{20} - 1}{11} \]
Now, we need to calculate \( 12^{20} - 1 \):
While the exact number \( 12^{20} \) is very large, we generally report sums like these as fractions or a combination of powers. For practical purposes, you would use computational tools to find \( 12^{20} \), and then perform the calculations.
In summary, the sum of the given geometric series \( \sum_{n=1}^{20} 12^{n-1} \) is:
\[ \frac{12^{20} - 1}{11} \]
To put this in a final box, we can state:
The sum is \( \frac{12^{20} - 1}{11} \) (exact value, can be computed for a numerical approximation).