Problem 27 - Thomas Calculus

Question

Evaluate the sums in Exercises \(19-28\). $$ \sum_{k=1}^{5} \frac{k^{3}}{225}+\left(\sum_{k=1}^{5} k\right)^{3} $$

Step-by-Step Solution

Answer

The evaluated sum is 3376.

Step 1: Understand the Expression

The expression given is the sum of two individual terms: \( \sum_{k=1}^{5} \frac{k^{3}}{225} \) and \( \left(\sum_{k=1}^{5} k\right)^{3} \). We need to evaluate each separately and then add the results together.

Step 2: Evaluate \( \sum_{k=1}^{5} \frac{k^3}{225} \)

To evaluate \( \sum_{k=1}^{5} \frac{k^3}{225} \), first find \( k^3 \) for \( k = 1 \) to \( k = 5 \):\[\begin{align*}1^3 &= 1, \2^3 &= 8, \3^3 &= 27, \4^3 &= 64, \5^3 &= 125.\end{align*}\]Next, calculate their sum: \( 1 + 8 + 27 + 64 + 125 = 225 \). Thus, the expression becomes \( \frac{225}{225} = 1 \).

Step 3: Evaluate \( \left(\sum_{k=1}^{5} k\right)^{3} \)

First, calculate \( \sum_{k=1}^{5} k \). The sum of the first five integers is: \( 1 + 2 + 3 + 4 + 5 = 15 \). Now, find its cube: \( 15^3 \). Calculate \( 15^3 \): \[15 imes 15 imes 15 = 3375\].

Step 4: Add the Results

Combine the results from Steps 2 and 3. We have already found \( \sum_{k=1}^{5} \frac{k^3}{225} = 1 \) and \( \left(\sum_{k=1}^{5} k \right)^3 = 3375 \). Add these values: \( 1 + 3375 = 3376 \).

Key Concepts

Arithmetic Series

An arithmetic series is a sequence of numbers in which the difference between consecutive terms is constant. In our exercise, the sequence of integers from 1 to 5 can be considered an arithmetic series. The sum of this series is calculated by adding up all the elements. For the given sequence (1+2+3+4+5), the sum is 15.

Key traits of arithmetic series include:

• Constant difference between terms.
• Sum represented by the formula: \( S_n = \frac{n}{2}(a + l) \), where \( n \) is the number of terms, \( a \) the first term, and \( l \) the last term.

This basic understanding helps us break down series into manageable problems, like in our exercise when dealing with the sum \( \sum_{k=1}^{5} k \). Knowing these properties, we computed the sum of this series quickly.

Cubic Sums

Cubic sums involve summing up cubes of a series of numbers. The exercise requires evaluating the cubic sum \( \sum_{k=1}^{5} k^3 \), where numbers from 1 to 5 are cubed and summed.

Features of cubic sums include:

• Cubing each term, i.e., raising it to the power of three.
• Adding these cubed values to find a total sum.

The method involves computing each cube, as shown: \( 1^3, 2^3, 3^3, 4^3, 5^3 \), resulting in \( 1, 8, 27, 64, 125 \) respectively. Adding these values gives 225, making the divisor \( \frac{225}{225} \) equal to 1 in this case. Understanding cubic sums aids in evaluating complex functions in calculus.

Mathematical Evaluation

Mathematical evaluation involves the step-by-step processing of expressions to obtain a numerical result. In our exercise, we evaluated two separate components to find their sum. Mathematical evaluation is crucial in breaking down problems into simpler parts, making complex calculations more approachable.

When solving the exercise:

• We first calculated the sum of cubic numbers divided by 225, yielding 1.
• Then, we determined the cubic power of the arithmetic series sum, resulting in 3375.
• Finally, these calculated values were summed to get the final answer of 3376.

This structured evaluation approach allows one to systematically tackle algebraic and calculus problems, ensuring clarity and precision.

Sigma Notation

Sigma notation, represented by the symbol \( \Sigma \), is a concise way to express sums. It's particularly useful in calculus and helps simplify the representation of series and sequences. In the exercise, \( \sum_{k=1}^{5} \frac{k^3}{225} + (\sum_{k=1}^{5} k)^3 \) showcases how sigma notation can encapsulate repetitive sum processes.

Key aspects of sigma notation are:

• Defines the start and end of a series with limits, here \( k = 1 \) to 5.
• Captures the expression for the terms being summed, like \( k^3 \) in the example.

Sigma notation simplifies complex calculations by reducing lengthy expressions into a more manageable form, helping you focus on evaluating the core elements of the series.