Hi, please could you show me the steps to calculate, in terms of k, the sum of 4(-2)^n  from n=1 up to k , thanks.

Certainly! To calculate the sum of the series \( S = \sum_{n=1}^{k} 4(-2)^n \), we will break down the process step by step.

### Step 1: Write the series explicitly
The sum can be written as:
\[
S = 4(-2)^1 + 4(-2)^2 + 4(-2)^3 + \cdots + 4(-2)^k
\]

Factoring out the 4 from the sum:
\[
S = 4\left[ (-2)^1 + (-2)^2 + (-2)^3 + \cdots + (-2)^k \right]
\]

This leaves us with the sum of powers of \(-2\), which is a **geometric series**.

### Step 2: Identify the geometric series
The geometric series is:
\[
(-2)^1 + (-2)^2 + (-2)^3 + \cdots + (-2)^k
\]
Here, the first term \( a = (-2)^1 = -2 \) and the common ratio \( r = -2 \).

The sum of the first \( k \) terms of a geometric series is given by the formula:
\[
\text{Sum of geometric series} = \frac{a(1 - r^k)}{1 - r}
\]

### Step 3: Apply the geometric series formula
For our series, \( a = -2 \), \( r = -2 \), and we are summing up to the \( k \)-th term, so the sum of the powers of \(-2\) becomes:
\[
\text{Sum} = \frac{-2(1 - (-2)^k)}{1 - (-2)}
\]
Simplifying the denominator:
\[
\text{Sum} = \frac{-2(1 - (-2)^k)}{3}
\]

### Step 4: Multiply by 4
Now, multiply the geometric sum by 4 (which we factored out earlier):
\[
S = 4 \times \frac{-2(1 - (-2)^k)}{3}
\]
Simplifying:
\[
S = \frac{-8(1 - (-2)^k)}{3}
\]

### Final Result:
Thus, the sum \( S \) is:
\[
S = \frac{-8(1 - (-2)^k)}{3}
\]

This is the sum of the series in terms of \( k \).

---

Do you have any questions or need further details?

Here are 5 related questions to consider:
1. How does the common ratio affect the behavior of a geometric series?
2. What happens to the sum as \( k \) increases in this series?
3. How would the sum change if the common ratio was a positive number?
4. Can you apply this method to sums where the base is a fraction?
5. How would you evaluate this sum for specific values of \( k \), like \( k = 3 \)?

**Tip:** Geometric series converge when the common ratio \( |r| < 1 \), but they grow rapidly otherwise, as seen in this case with \( r = -2 \).

Hi, please could you show me the steps to calculate, in terms of k, the sum of 4(-2)^n from n=1 up to k, thanks.

Calculate the sum of the geometric series 4(-2)^n from n=1 to k with a step-by-step breakdown. This problem involves key concepts in geometric series, exponents, and summation, suitable for students in Grades 10-12.

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