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The problem consists of two parts:

1. \( (3)^2 + (3)^3 + (3)^4 + (3)^6 \)
2. \( \sum_{j=1}^{5} \left( \frac{-1}{4} \right)^j \)

### Solution for Part 1:
We will compute each power of 3 individually and then sum the terms:
- \( (3)^2 = 9 \)
- \( (3)^3 = 27 \)
- \( (3)^4 = 81 \)
- \( (3)^6 = 729 \)

Now summing them up:
\[
9 + 27 + 81 + 729 = 846
\]

### Solution for Part 2:
We will evaluate the sum \( \sum_{j=1}^{5} \left( \frac{-1}{4} \right)^j \). This involves calculating the first 5 terms of the geometric series:
- \( \left( \frac{-1}{4} \right)^1 = \frac{-1}{4} \)
- \( \left( \frac{-1}{4} \right)^2 = \frac{1}{16} \)
- \( \left( \frac{-1}{4} \right)^3 = \frac{-1}{64} \)
- \( \left( \frac{-1}{4} \right)^4 = \frac{1}{256} \)
- \( \left( \frac{-1}{4} \right)^5 = \frac{-1}{1024} \)

Now summing these terms:
\[
\frac{-1}{4} + \frac{1}{16} + \frac{-1}{64} + \frac{1}{256} + \frac{-1}{1024}
\]

To sum fractions, we first get a common denominator (1024):
\[
\frac{-1}{4} = \frac{-256}{1024}, \quad \frac{1}{16} = \frac{64}{1024}, \quad \frac{-1}{64} = \frac{-16}{1024}, \quad \frac{1}{256} = \frac{4}{1024}, \quad \frac{-1}{1024} = \frac{-1}{1024}
\]

Now summing them:
\[
\frac{-256 + 64 - 16 + 4 - 1}{1024} = \frac{-205}{1024}
\]

Thus, the sum of the series is \( \frac{-205}{1024} \).

### Final Answer:
- Part 1: \( 846 \)
- Part 2: \( \frac{-205}{1024} \)

Would you like further details on any part of the solution?

Here are 5 related questions:
1. How do you simplify geometric series in general?
2. Can you explain how to find the sum of a geometric series using a formula?
3. What happens if we change the base in exponential sums?
4. How do fractional powers affect geometric series sums?
5. Can you demonstrate another example with different exponents and terms?

**Tip:** Always ensure the fractions are expressed with a common denominator before summing them up.

Compute each sum below. If applicable, write your answer as a fraction. The given expression includes two parts: (3)^2 + (3)^3 + (3)^4 + (3)^6 and the sum Σ from j=1 to 5 of (-1/4)^j.

Learn how to compute the sum of the series involving powers of 3 and a geometric series with fractional terms. This solution includes step-by-step instructions for solving (3)^2 + (3)^3 + (3)^4 + (3)^6 and a summation of (-1/4)^j for j = 1 to 5.

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