Answer:
The given series is a geometric series with a common ratio of \( \frac{3}{4} \) because each term is obtained by multiplying the previous term by \( \frac{3}{4} \).
The general form of a geometric series is \( a + ar + ar^2 + \ldots + ar^{n-1} \), where:
- \( a \) is the first term,
- \( r \) is the common ratio,
- \( n \) is the number of terms.
In this case:
- \( a = 3 \) (the first term),
- \( r = \frac{3}{4} \) (the common ratio),
- \( n = 8 \) (the number of terms).
The sum of the first \( n \) terms of a geometric series is given by the formula \( S_n = \frac{a(1 - r^n)}{1 - r} \).
Let's calculate it for the given series:
\[ S_8 = \frac{3(1 - \left(\frac{3}{4}\right)^8)}{1 - \frac{3}{4}} \]
\[ S_8 = \frac{3(1 - \frac{6561}{65536})}{\frac{1}{4}} \]
\[ S_8 = \frac{3(\frac{58975}{65536})}{\frac{1}{4}} \]
\[ S_8 = \frac{176925}{8192} \]
Now, to the nearest integer, \( S_8 \approx 22 \).
So, the sum of the first 8 terms of the given series is approximately 22 (rounded to the nearest integer).