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

The given sum is:

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

### Step 1: Factor out the constant
We can factor out the constant 40 from the summation:

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

### Step 2: Change index of summation
Let's shift the index by 1 to simplify the powers of \( 0.95 \). If we let \( m = n+1 \), then:

\[
S = 40 \cdot \sum_{m=2}^{42} (0.95)^m
\]

We can rewrite this as:

\[
S = 40 \cdot \left( \sum_{m=1}^{42} (0.95)^m - (0.95)^1 \right)
\]

### Step 3: Use the formula for the sum of a geometric series
The sum of a geometric series is given by:

\[
\sum_{m=1}^{N} r^m = \frac{r(1 - r^N)}{1 - r}
\]

In this case, \( r = 0.95 \), and we sum up to \( m = 42 \). So we calculate:

\[
S = 40 \cdot \left( \frac{0.95(1 - (0.95)^{42})}{1 - 0.95} - 0.95 \right)
\]

Let me calculate the value of this sum.The value of the given expression, rounded to the nearest integer, is \( 634 \).

Do you need any further details or clarification?

Here are some related questions:
1. How does the geometric series formula apply to this problem?
2. What is the effect of changing the ratio \( 0.95 \) in the sum?
3. What if the upper limit of summation changes to 50 instead of 41?
4. How does the result change if the constant multiplier 40 changes?
5. How would you compute such sums for other values of \( r \)?

**Tip**: When dealing with geometric series, always check if the common ratio \( r \) is less than 1, as it simplifies the summation considerably.

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

Learn how to calculate the sum of a geometric series for the expression 40(0.95)^(n+1), with a detailed solution and the application of the geometric series sum formula. Suitable for high school students, this problem covers algebraic techniques and exponential functions.

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