Problem 15 - Calculus: Early Transcendental Functions

Question

Use summation rules to compute the sum. $$\sum_{i=1}^{200}\left(4-3 i-i^{2}\right)$$

Step-by-Step Solution

Answer

The sum is \( -270804500 \).

Step 1: Split the Sum

The given summation can be split into three different sums: \( \sum_{i=1}^{200}\left(4-3 i-i^{2}\right) = \sum_{i=1}^{200}4 - \sum_{i=1}^{200}3i - \sum_{i=1}^{200}i^{2} \)

Step 2: Apply the Summation Rules

The first term, \(\sum_{i=1}^{200}4\), simplifies to \( 4*200 = 800 \) because the sum of \( n \) identical terms is the term times \( n \). The second term, \(\sum_{i=1}^{200}3i\), uses the sum formula for consecutive integers: \( \sum_{i=1}^{n} ai = a*(n*(n+1))/2 \). Hence, the summation simplifies to \( 3*(200 * 201) / 2 = 60300 \). The third term, \(\sum_{i=1}^{n} i^{2}\), uses the sum formula for squares of integers: \(\sum_{i=1}^{n} i^{2} = n*(n+1)*(2n+1)/6\). The third term therefore simplifies to \(200 * 201 * 401 / 6 = 27027000 \).

Step 3: Compute the Final Sum

Combining all the results from Step 2, the final computed sum is \( 800 - 60300 - 27027000 \).

Key Concepts

Sum of Consecutive Integers

Finding the sum of consecutive integers is a common task in mathematics. This involves summing up a sequence of numbers like 1, 2, 3, and so on, up to a certain number. The formula used to calculate the sum of the first n consecutive integers is:

• \( \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \)

In this formula, \( n \) represents the last integer in the series. This equation is simple but powerful. It allows you to quickly find the sum without having to add each integer one by one.
For example, using the formula for \( n = 200 \), the sum is \( \frac{200(200+1)}{2} = 20100 \). This tells us that summing all numbers from 1 to 200 results in 20100. By using this formula, you can save time and ensure accuracy when dealing with long lists of numbers.

Sum of Squares of Integers

The sum of squares of integers is another useful formula. It involves summing the squares of consecutive numbers, like \( 1^2, 2^2, 3^2, \) and so on, up to \( n^2 \). The formula for the sum of squares is:

• \( \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \)

Here, \( n \) is the highest number in the sequence. This formula is particularly handy in calculating the sum efficiently, especially when \( n \) is a large number.
For example, if \( n = 200 \), then the sum of squares is \( \frac{200(200+1)(2*200+1)}{6} = 2702700 \).
This sum represents the total when you square each number from 1 to 200 and add them. Such calculations are critical in various areas of mathematics, including statistics and physics.

Splitting Summations

Splitting summations is a mathematical technique used to simplify complex expressions by breaking them down into smaller parts. When you have a sum that combines different terms, you can separate each term and sum them individually. This approach uses basic algebra to handle summation problems that involve multiple expressions.
Take the example \( \sum_{i=1}^{200}(4 - 3i - i^2) \). You can split this into three separate sums:

• \( \sum_{i=1}^{200} 4 \)
• \( -\sum_{i=1}^{200} 3i \)
• \( -\sum_{i=1}^{200} i^2 \)

By splitting it up, you can apply specific summation formulas to each part. The sum of a constant is simple, as seen in \( \sum_{i=1}^{200} 4 = 4*200 = 800 \). For terms like \( -\sum_{i=1}^{200} 3i \), you can use the sum of consecutive integers formula, and for \( -\sum_{i=1}^{200} i^2 \), apply the sum of squares formula.
Once these individual sums are calculated, combine them to get the final result. This strategy of breaking down helps manage and solve complex summation problems more effectively.