The problem asks to find the summation up to the 8th term of the given sequence: $3, \_, 27, 81, \ldots$.

Let's first identify the sequence.

The given terms are $3, \_, 27, 81$. This suggests that the sequence is geometric, as each term is increasing by a constant ratio.

1. To confirm, divide the third term by the first term: $\frac{27}{3} = 9$ So, the common ratio $r$ is $3$, because the missing second term should be $9$ (since $3 \times 3 = 9$).

Thus, the sequence is $3, 9, 27, 81, \ldots$, with the first term $a = 3$ and the common ratio $r = 3$.

Finding the sum of the first 8 terms of a geometric series:

The sum $S_n$ of the first $n$ terms of a geometric series is given by the formula: $S_n = \frac{a(1 - r^n)}{1 - r}$ where:

• $a = 3$ (first term),
• $r = 3$ (common ratio),
• $n = 8$ (number of terms).

Now, substituting the values: $S_8 = \frac{3(1 - 3^8)}{1 - 3}$ $S_8 = \frac{3(1 - 6561)}{-2}$ $S_8 = \frac{3 \times (-6560)}{-2}$ $S_8 = \frac{-19680}{-2} = 9840$

Thus, the sum of the first 8 terms of the sequence is 9840.

Would you like a detailed breakdown of the steps or have any questions?

Here are some related questions:

1. How do you find the nth term of a geometric sequence?
2. What happens if the common ratio is less than 1?
3. How does the formula for the sum of infinite geometric series differ from that of finite ones?
4. What are the applications of geometric sequences in real-world problems?
5. How would the formula change if the common ratio was negative?

Tip: Always check the common ratio between terms to identify a geometric sequence easily.