Problem 19 - University Calculus: Early Transcendentals

Question

Evaluate the sums. $$\begin{array}{lll} \text { a. } \sum_{k=1}^{10} k & \text { b. } \sum_{k=1}^{10} k^{2} & \text { c. } \sum_{k=1}^{10} k^{3} \end{array}$$

Step-by-Step Solution

Answer

a. 55, b. 385, c. 3025

Step 1: Understanding the Problem

We are given three different sums to evaluate: \( \sum_{k=1}^{10} k \), \( \sum_{k=1}^{10} k^2 \), and \( \sum_{k=1}^{10} k^3 \). Each of these expressions is a summation notation, which indicates that we are summing a sequence of numbers from \( k = 1 \) to \( k = 10 \). Our goal is to compute the total sum for each of these functions, one linear, one quadratic, and one cubic.

Step 2: Evaluate the Linear Sum

To evaluate \( \sum_{k=1}^{10} k \), we recognize this as the sum of the first 10 natural numbers. Using the formula for the sum of the first \( n \) natural numbers, \( \frac{n(n+1)}{2} \), we substitute \( n = 10 \): \[ \sum_{k=1}^{10} k = \frac{10 \times 11}{2} = 55. \]

Step 3: Evaluate the Quadratic Sum

For \( \sum_{k=1}^{10} k^2 \), we apply the formula for the sum of squares of the first \( n \) natural numbers, which is \( \frac{n(n+1)(2n+1)}{6} \). Substituting \( n = 10 \), we have: \[ \sum_{k=1}^{10} k^2 = \frac{10 \times 11 \times 21}{6} = 385. \]

Step 4: Evaluate the Cubic Sum

To evaluate \( \sum_{k=1}^{10} k^3 \), we use the formula for the sum of cubes of the first \( n \) natural numbers, \( \left( \frac{n(n+1)}{2} \right)^2 \). For \( n = 10 \), this becomes: \[ \sum_{k=1}^{10} k^3 = \left( \frac{10 \times 11}{2} \right)^2 = 3025. \]

Step 5: Verify Results and Conclude

We've calculated the sums based on their respective formulas: \( \sum_{k=1}^{10} k = 55 \), \( \sum_{k=1}^{10} k^2 = 385 \), and \( \sum_{k=1}^{10} k^3 = 3025 \). Ensure these results match the expected values from standard formulas typically used at this level.

Key Concepts

Understanding Natural Numbers

The natural numbers are the simplest set of numbers that we use for counting. These numbers start from 1 and extend indefinitely to numbers like 2, 3, 4, and so on. They are often represented in mathematical notation as \( \{1, 2, 3, \ldots\} \). Natural numbers have no fractions or negative values, making them very straightforward to work with.

In mathematics, natural numbers form the basis for many types of calculations, including summations. When talking about summations over natural numbers, it typically means adding them in sequence. This can help solve real-life counting problems or discover patterns within the numbers.

So, if ever you need to tally up items or count steps, natural numbers are the go-to set for such tasks.

Exploring the Sum of Squares

The sum of squares refers to adding up the squares of a series of numbers. For example, in the case \( \sum_{k=1}^{10} k^2 \), it means 1 squared plus 2 squared, all the way up to 10 squared. Each number within the sequence is squared first and then summed, which often finds its use in statistical calculations and geometry.

Mathematically, this is not as simple as a straight sum, and specific formulas come in handy for large numbers. The formula to find the sum of squares for the first \( n \) natural numbers is \( \frac{n(n+1)(2n+1)}{6} \).

Why is this useful? It's because it simplifies computation for higher numbers without the need to compute each number individually. This method is often essential when calculating variances in data sets in statistics, showing how values deviate from the mean.

Employing such formulas simplifies tasks that would otherwise require countless repetitive steps.

Understanding the Sum of Cubes

In mathematics, the sum of cubes refers to adding the cube of numbers in a sequence. For any integer sequence, you calculate each number raised to the power of 3 and then find their summation. This concept is mathematically denoted as \( \sum_{k=1}^{n} k^3 \).

Finding the sum of cubes is often more complex than regular addition or even the sum of squares. Luckily, mathematics provides a simplified formula: \( \left(\frac{n(n+1)}{2}\right)^2 \). This not only makes computations faster but also unveils interesting results, such as realizing that the sum of cubes sequence forms perfect squares. Hence for our example using \( n = 10 \), the sum of cubes becomes 3025.

Application of such formulas is vital in many areas, particularly in algebra and number theory, as they simplify complex problems. Additionally, they reveal curious properties of numbers, for instance, that these cubes can produce square numbers.

Decoding Mathematical Formulas

Mathematical formulas are the backbone of mathematics. They offer compact ways to represent complex operations through symbols and numbers. Formulas quickly communicate relationships and functions between different components in a mathematical expression.

With each formula, solving problems becomes far more manageable. In our exercises, we used arithmetic formulas to find the sum of natural numbers, squares, and cubes. Each formula provides an efficient means for computation, reducing the time required to carry out long series of additions manually.

Here is how they work:

• The sum of natural numbers formula: \( \frac{n(n+1)}{2} \)
• The sum of squares formula: \( \frac{n(n+1)(2n+1)}{6} \)
• The sum of cubes formula: \( \left(\frac{n(n+1)}{2}\right)^2 \)

These formulas simplify complex tasks, proving invaluable in both mathematical studies and real-world applications. Recognizing when to use specific formulas makes problem-solving systematic and systematic.