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.