Find the value of the following expression and round to the nearest integer:
sum, from, n, equals, 1, to, 41, of, 40, left parenthesis, 0, point, 9, 5, right parenthesis, start superscript, n, plus, 1, end superscript
n=1
∑
41
​
40(0.95)
n+1

We are asked to find the value of the following sum:

\[
S = \sum_{n=1}^{41} 40 \cdot (0.95)^{n+1}
\]

### Step-by-step solution:

1. **Factor out the constant**: The constant \( 40 \) can be factored out of the summation:
   \[
   S = 40 \cdot \sum_{n=1}^{41} (0.95)^{n+1}
   \]
   This is equivalent to:
   \[
   S = 40 \cdot \sum_{n=1}^{41} (0.95)^{n+1}
   \]

2. **Rewrite the exponent**: Let’s adjust the limits and rewrite the exponent:
   \[
   S = 40 \cdot \sum_{n=1}^{41} (0.95)^{n+1} = 40 \cdot \sum_{n=2}^{42} (0.95)^n
   \]

3. **Summation of a geometric series**: The sum we now have is a geometric series:
   \[
   \sum_{n=2}^{42} (0.95)^n
   \]
   A geometric series has the form:
   \[
   \sum_{n=a}^{b} r^n = \frac{r^a(1 - r^{b-a+1})}{1 - r}
   \]
   Here \( r = 0.95 \), \( a = 2 \), and \( b = 42 \). Plugging in these values, we can calculate the sum.

Let's calculate this step by step.The value of the expression is approximately \( 633.85 \), which rounds to \( 634 \) when rounded to the nearest integer.

Do you want more details on this calculation or have any questions?

Here are 5 related questions to further explore this topic:
1. How do you derive the formula for the sum of a geometric series?
2. What happens to the sum if the ratio \( r \) is greater than 1?
3. How would the result change if we extended the summation beyond \( n = 41 \)?
4. Can this method be used for sums with different constants or ratios?
5. How would you calculate a geometric series if the ratio is negative?

**Tip:** The geometric series converges when the ratio \( |r| < 1 \), leading to finite sums for infinite series.

Find the value of the following expression and round to the nearest integer: sum from n equals 1 to 41 of 40 * (0.95)^(n + 1)

Learn how to calculate the sum of the geometric series from n = 1 to 41 for the expression 40 * (0.95)^(n + 1), using the geometric series formula. This step-by-step solution simplifies the summation and explores key mathematical concepts such as exponents and series. Suitable for students in Grades 10-12.

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