Let's solve these problems step-by-step.

Problem 1:

The first part is: $2 + 2(-2) + 2(-2)^2 + \dots + 2(-2)^6$

This is a geometric sequence where:

• The first term $a = 2$,
• The common ratio $r = -2$,
• The number of terms $n = 7$ (from $2(-2)^0$ to $2(-2)^6$).

The formula for the sum of the first $n$ terms of a geometric sequence is: $S_n = a \frac{1 - r^n}{1 - r}$

Plugging in the values: $S_7 = 2 \cdot \frac{1 - (-2)^7}{1 - (-2)} = 2 \cdot \frac{1 + 128}{1 + 2} = 2 \cdot \frac{129}{3} = 2 \cdot 43 = 86$

So, the sum is: $\boxed{86}$

Problem 2:

The second part is: $\sum_{k=1}^9 \left( \frac{1}{3} \right)^k$

This is also a geometric sequence where:

• The first term $a = \frac{1}{3}$,
• The common ratio $r = \frac{1}{3}$,
• The number of terms $n = 9$.

Using the sum formula: $S_n = a \frac{1 - r^n}{1 - r}$

Substitute the values: $S_9 = \frac{1}{3} \cdot \frac{1 - \left( \frac{1}{3} \right)^9}{1 - \frac{1}{3}}$

Simplifying the denominator: $1 - \frac{1}{3} = \frac{2}{3}$

Thus: $S_9 = \frac{1}{3} \cdot \frac{1 - \frac{1}{3^9}}{\frac{2}{3}} = \frac{1}{3} \cdot \frac{3}{2} \left(1 - \frac{1}{3^9}\right) = \frac{1}{2} \left(1 - \frac{1}{19683}\right)$

This simplifies to: $S_9 = \frac{1}{2} \left(\frac{19682}{19683}\right) = \frac{19682}{39366}$

Thus, the answer as a fraction is: $\boxed{\frac{19682}{39366}}$

Would you like more details on any step, or do you have questions about these solutions?

Here are some related questions to explore:

1. How do we determine if a sequence is geometric?
2. What happens to the sum of a geometric series as $n$ approaches infinity if $|r| < 1$?
3. How does the common ratio affect the behavior of a geometric sequence?
4. Can geometric series be used to approximate certain types of functions?
5. How does the formula for the sum change if the common ratio $r = 1$?

Tip: When dealing with fractions in geometric series, simplifying the result at each step can help avoid errors in large calculations.