Let's analyze and solve the two problems displayed in the image.

1st Problem:

$(-3) + (-3)^2 + (-3)^3 + \cdots + (-3)^8$

This is a series of powers of $-3$, starting from $(-3)^1$ up to $(-3)^8$. We can calculate each term individually and sum them:

• $(-3)^1 = -3$
• $(-3)^2 = 9$
• $(-3)^3 = -27$
• $(-3)^4 = 81$
• $(-3)^5 = -243$
• $(-3)^6 = 729$
• $(-3)^7 = -2187$
• $(-3)^8 = 6561$

Now, summing them up: $S = -3 + 9 - 27 + 81 - 243 + 729 - 2187 + 6561$ $S = 6920$

2nd Problem:

$\sum_{j=1}^{9} 5 \left( \frac{1}{4} \right)^j$

This is a geometric series where the first term $a = 5 \cdot \left( \frac{1}{4} \right) = \frac{5}{4}$ and the common ratio $r = \frac{1}{4}$. The number of terms is $9$.

The formula for the sum of a geometric series is: $S_n = \frac{a(1 - r^n)}{1 - r}$ Substituting the values: $S_9 = \frac{\frac{5}{4} \left(1 - \left(\frac{1}{4}\right)^9 \right)}{1 - \frac{1}{4}}$ $S_9 = \frac{\frac{5}{4} \left(1 - \frac{1}{262144} \right)}{\frac{3}{4}}$ $S_9 = \frac{5}{4} \times \frac{262143}{262144} \times \frac{4}{3}$ Simplifying: $S_9 = \frac{5 \times 262143}{3 \times 262144} = \frac{1310715}{786432}$ Thus, the exact value of the sum is $\frac{1310715}{786432}$, which can be simplified further if required.

Let me know if you'd like a more detailed explanation or have any questions!

Related Questions:

1. How do you identify a geometric series and its properties?
2. What is the general formula for summing powers of integers?
3. Can geometric series have negative terms? How does that affect the sum?
4. How does the sum formula for a geometric series change if the ratio is greater than 1?
5. How do you simplify large fractions like $\frac{1310715}{786432}$?

Tip:

When dealing with powers of negative numbers, check the exponent's parity. Even exponents yield positive results, while odd exponents yield negative results.