Find the value of the following expression and round to the nearest integer:
sum, from, n, equals, 2, to, 28, of, 10, left parenthesis, 1, point, 3, right parenthesis, start superscript, n, plus, 1, end superscript
n=2
∑
28
​
10(1.3)
n+1

We are tasked with evaluating the following sum:

\[
S = \sum_{n=2}^{28} 10 \times (1.3)^{n+1}
\]

### Step-by-Step Solution:

1. **Expand the sum:**

   The summation means we need to calculate the expression for each \( n \) from \( n=2 \) to \( n=28 \) and sum the values.

   \[
   S = 10 \times (1.3)^3 + 10 \times (1.3)^4 + \cdots + 10 \times (1.3)^{29}
   \]

2. **Factor out the constant 10:**

   \[
   S = 10 \times \left( (1.3)^3 + (1.3)^4 + \cdots + (1.3)^{29} \right)
   \]

3. **Sum of a geometric series:**

   The expression inside the parentheses is a geometric series where the first term is \( a = (1.3)^3 \), the common ratio is \( r = 1.3 \), and the number of terms is \( 29 - 3 + 1 = 27 \).

   The sum of a geometric series is given by:

   \[
   \text{Sum} = \frac{a(r^n - 1)}{r - 1}
   \]

   where:
   - \( a = (1.3)^3 \),
   - \( r = 1.3 \),
   - \( n = 27 \).

   Let's calculate the sum now.

The value of the given sum, rounded to the nearest integer, is \( 87,260 \).

Would you like further details or have any questions?

Here are 5 related questions to expand on this:

1. How is the formula for the sum of a geometric series derived?
2. What happens if the common ratio in a geometric series is less than 1?
3. How does the sum of a geometric series change if the number of terms increases?
4. Can this approach be applied to series with varying coefficients? Why or why not?
5. How does changing the base (1.3) affect the growth rate of the series?

**Tip:** In geometric series, the common ratio determines whether the series grows, decays, or stays constant.

Find the value of the following expression and round to the nearest integer: sum from n = 2 to 28 of 10(1.3)^(n+1)

Learn how to evaluate the sum of the geometric series from n=2 to n=28 for the expression 10(1.3)^(n+1). This solution demonstrates step-by-step calculations using the geometric series sum formula and exponential growth. Suitable for Grades 9-12 students.

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