Problem 76
Question
Find the partial sum without using a graphing utility. $$\sum_{n=1}^{250}(1000-n)$$
Step-by-Step Solution
Verified Answer
The partial sum of the series is \(218625\).
1Step 1: Identify the type of series
The given series, \(\sum_{n=1}^{250}(1000-n)\), is an arithmetic series where each term is a fixed difference away from the next term. In this case, the common difference is -1.
2Step 2: Calculate the first term (a) and the last term (l)
To apply the formula for the sum of an arithmetic series, we first need to calculate the first term (a) and the last term (l). The first term is when \(n=1\), so \(a = 1000 - 1 = 999\). The last term is when \(n=250\), so \(l = 1000 - 250 = 750\).
3Step 3: Apply the formula for sum of an arithmetic series
The formula for the sum of an arithmetic series is \(\frac{n}{2}*(a + l)\), where n is the number of terms, a is the first term, and l is the last term. Applying these values to the formula: The number of terms, \(n = 250\). The first term, a = 999. The last term, l = 750. Substituting these into the formula, we get \(\frac{250}{2} * (999 + 750) = 125 * 1749\).
Key Concepts
Partial Sum CalculationArithmetic ProgressionSum of Series Formula
Partial Sum Calculation
Understanding how to calculate the partial sum of a series is crucial for comprehending more complex concepts in mathematics. A partial sum is simply the sum of a portion of a sequence of numbers. In the context of an arithmetic series, we're dealing with a sequence where each number differs from the previous one by a constant amount—a property known as the common difference.
To find the partial sum (which is often denoted as Sn), you need to know the first term, the last term, and the total number of terms you wish to add. In the given exercise, \( \sum_{n=1}^{250}(1000-n) \), we needed to find the first and the last term in order to plug these into the sum formula. Here's an easy-to-remember method:
To find the partial sum (which is often denoted as Sn), you need to know the first term, the last term, and the total number of terms you wish to add. In the given exercise, \( \sum_{n=1}^{250}(1000-n) \), we needed to find the first and the last term in order to plug these into the sum formula. Here's an easy-to-remember method:
- First Term (a): Set n=1 and calculate the term
- Last Term (l) or Termn: Calculate the term using the last number of n in the series
- Number of Terms (n): This is often given or can be calculated as (Termn - a)/common difference + 1
Arithmetic Progression
An arithmetic progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a constant to the previous term. This constant is known as the common difference (d). The concept of an arithmetic progression makes it straightforward to predict subsequent numbers in a sequence once the first term and the common difference are known.
For instance, if we have an AP starting with 3 and a common difference of 2, the sequence looks like this: 3, 5, 7, 9, and so on. In our exercise, our AP is unusual because the common difference is negative, which means the series is decreasing. This is still an arithmetic progression, though! Understanding the nature of the AP is essential to correctly apply formulas and find sums efficiently, whether it is increasing or decreasing. Arithmetic progressions are everywhere, from counting days to managing finances. The pattern they provide offers a rhythm to the numbers that is both predictable and immensely useful.
For instance, if we have an AP starting with 3 and a common difference of 2, the sequence looks like this: 3, 5, 7, 9, and so on. In our exercise, our AP is unusual because the common difference is negative, which means the series is decreasing. This is still an arithmetic progression, though! Understanding the nature of the AP is essential to correctly apply formulas and find sums efficiently, whether it is increasing or decreasing. Arithmetic progressions are everywhere, from counting days to managing finances. The pattern they provide offers a rhythm to the numbers that is both predictable and immensely useful.
Sum of Series Formula
A sum of series formula is a mathematical way to add all numbers in a series quickly, without needing to add each number individually. For arithmetic series, the formula is particularly elegant and takes the form \( \frac{n}{2}*(a + l) \), where n is the number of terms, a is the first term, and l is the last term.
Let's use our example from earlier: \( \frac{250}{2} * (999 + 750) \). Why does this formula work? It's based on the concept that the sum of an arithmetic series is equivalent to the average of the first and last terms, multiplied by the number of terms. If you think about it, this makes sense because each pair of terms equidistant from the start and end of the series will add up to the same value, essentially forming pairs that average to the same number.
Let's use our example from earlier: \( \frac{250}{2} * (999 + 750) \). Why does this formula work? It's based on the concept that the sum of an arithmetic series is equivalent to the average of the first and last terms, multiplied by the number of terms. If you think about it, this makes sense because each pair of terms equidistant from the start and end of the series will add up to the same value, essentially forming pairs that average to the same number.
Pro Tip:
When you're feeling stuck, remember that an arithmetic series is just a list of numbers with a regular interval between them. So, when you use the sum of series formula, you're not just crunching numbers – you're harnessing the power of patterns to make the complex simple!Other exercises in this chapter
Problem 75
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