Problem 28 - Thomas Calculus

Question

Evaluate the sums in Exercises \(19-28\). $$ \left(\sum_{k=1}^{7} k\right)^{2}-\sum_{k=1}^{7} \frac{k^{3}}{4} $$

Step-by-Step Solution

Answer

The evaluated sum is 588.

Step 1: Calculate the sum of integers

First, calculate \( \sum_{k=1}^{7} k \), which is the sum of the first 7 integers. The sum can be found using the formula for the sum of an arithmetic series: \[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \]For \( n = 7 \), the sum is:\[ \sum_{k=1}^{7} k = \frac{7(8)}{2} = 28 \]

Step 2: Square the sum of integers

Next, square the result from Step 1:\[ (\sum_{k=1}^{7} k)^{2} = 28^{2} = 784 \]

Step 3: Compute the sum of cubes divided by 4

Now, calculate \( \sum_{k=1}^{7} \frac{k^{3}}{4} \). First, find \( \sum_{k=1}^{7} k^{3} \). The sum of cubes is given by:\[ \sum_{k=1}^{n} k^{3} = \left(\frac{n(n+1)}{2}\right)^{2} \]For \( n = 7 \),\[ \sum_{k=1}^{7} k^{3} = \left(\frac{7 \times 8}{2}\right)^{2} = 28^{2} = 784 \]Now divide by 4:\[ \sum_{k=1}^{7} \frac{k^{3}}{4} = \frac{784}{4} = 196 \]

Step 4: Subtract the results from previous steps

Subtract the result from Step 3 from the result in Step 2:\[ (\sum_{k=1}^{7} k)^{2} - \sum_{k=1}^{7} \frac{k^{3}}{4} = 784 - 196 = 588 \]

Key Concepts

Arithmetic Series

An arithmetic series refers to the sum of a sequence of numbers in which each term after the first is formed by adding a constant to the previous term. This constant is called the common difference. For example, in the sequence 1, 2, 3, 4,... the common difference is 1.
The formula to find the sum of the numbers in an arithmetic sequence up to any number \( n \) is given by \[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \].
This is a handy formula to quickly calculate the sum of a series of consecutive numbers up to 7, as seen in the problem. Here, you can see that \( \sum_{k=1}^{7} k = \frac{7 \times 8}{2} = 28 \).

• This helps in understanding patterns within sequences of numbers.
• It's foundational in studying more complex mathematical concepts.

Sum of Cubes

The sum of cubes is an interesting and useful mathematical concept. It involves adding together the cubes of a sequence of integers. For each integer \( k \), you calculate its cube, \( k^3 \), and then sum all of these values. This is expressed as \( \sum_{k=1}^{n} k^3 \).
Interestingly, the sum of cubes of the first \( n \) natural numbers is equal to \( \left(\frac{n(n+1)}{2}\right)^2 \).
This formula reflects a deep relationship between the sequence of numbers and their cubes, showing how structured these sums are. In our exercise, for \( n = 7 \): \( \sum_{k=1}^{7} k^3 = 28^2 = 784 \).

• Using this helps simplify complex calculations.
• Makes it easier to compute larger series accurately.

Summation Notation

Summation notation is a concise way to represent the sum of a sequence of terms. It utilizes the Greek letter sigma (\( \Sigma \)) to indicate summation.
For example, \( \sum_{k=1}^{7} k \) means "sum from \( k = 1 \) to \( k = 7 \)". Each term of the sequence is added together. This allows mathematicians to present complex series calculations in a simplified form.
Using summation helps in:

• Providing a standardized method to denote large sums.
• Helping to apply mathematical formulas directly without the need for writing out every single term.

The use of \( \sum \) in the problem helps us to compactly express the operations needed to find both the sum of integers and their cubes.

Step-by-Step Calculation

Breaking a problem down into individual steps is essential for understanding and solving any mathematical exercise. It involves addressing the problem one piece at a time.
This method ensures that each aspect is properly managed and that errors are spotted more easily. In our own example:

• We start by calculating the arithmetic series sum.
• Then proceed to square the result to handle the next part of the problem.
• Compute the sum of cubes and divide as required next.
• Finally, subtract the two results to arrive at the answer.

By tackling each step separately, we create a clear path towards the solution. This technique allows anyone learning calculus to not only solve the problem at hand but to also practice analytical and logical thinking, which are key skills in mathematics.