Chapter 9: Problem 5
Evaluate: a. \(\sum_{k=1}^{100} k\) b. \(\sum_{k=3}^{100} k\) c. \(\sum_{k=18}^{100} k \quad\) d. \(\quad \sum_{k=1}^{100} 2 k\) e. \(\sum_{k=1}^{10} k^{2}\) f. \(\sum_{k=3}^{100} k^{3}\) g. \(\quad \sum_{k=1}^{100}(k+5)\) h. \(\sum_{k=50}^{100}(3 k+7)\) i. \(\quad \sum_{k=1}^{10} 5 k^{2} \quad\) j. \(\quad \sum_{k=3}^{100}\left(k^{3}-9\right)\) k. \(\sum_{k=32}^{78}\left(3 k^{2}-k\right) \quad\) l. \(\quad\left(\sum_{k=50}^{100} 3 k\right)+7\)
Short Answer
Step by step solution
Evaluate \( \sum_{k=1}^{100} k \)
Evaluate \( \sum_{k=3}^{100} k \)
Evaluate \( \sum_{k=18}^{100} k \)
Evaluate \( \sum_{k=1}^{100} 2k \)
Evaluate \( \sum_{k=1}^{10} k^2 \)
Evaluate \( \sum_{k=3}^{100} k^3 \)
Evaluate \( \sum_{k=1}^{100} (k+5) \)
Evaluate \( \sum_{k=50}^{100} (3k+7) \)
Evaluate \( \sum_{k=1}^{10} 5k^2 \)
Evaluate \( \sum_{k=3}^{100} (k^3 - 9) \)
Evaluate \( \sum_{k=32}^{78} (3k^2 - k) \)
Evaluate \( \left(\sum_{k=50}^{100} 3k\right) + 7 \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Arithmetic Series
One of the easiest examples is the sum of the first \( n \) positive integers. The formula to calculate this is \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \). This formula allows us to quickly find the sum without needing to add each number manually.
When calculating the sum over a specified range, such as from some integer \( a \) to another integer \( b \), the sum can be found by subtracting the sum of numbers below \( a \) from the sum of numbers up to \( b \). For example, \( \sum_{k=3}^{100} k \) is found by \( \sum_{k=1}^{100} k - \sum_{k=1}^{2} k \). This approach simplifies the process, especially for large sections of an arithmetic series.
Sum of Squares
\[ \sum_{k=1}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6} \]
This formula helps you quickly compute the result without calculating each square separately.
For example, if you want to calculate \( \sum_{k=1}^{10} k^2 \), you can simply plug \( n=10 \) into the formula, resulting in \( \frac{10 \times 11 \times 21}{6} = 385 \). This is far more efficient than manually calculating \( 1^2 + 2^2 + ... + 10^2 \).
The sum of squares is particularly useful in fields like statistics for calculating variance and standard deviation.
Sum of Cubes
\[ \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 \]
This formula tells us that the sum of the cubes of the first \( n \) natural numbers is actually the square of the sum of those numbers.
To illustrate, let's calculate \( \sum_{k=1}^{100} k^3 \). First, you calculate \( \frac{100 \times 101}{2} = 5050 \), then square 5050 to get \( 5050^2 = 25502500 \).
This method provides a powerful shortcut in summing cubes and reveals a beautiful mathematical symmetry, linking it back to basic arithmetic series.
Mathematical Formulas
They serve as shortcuts, reducing the need for repeated addition or multiplication by providing a more efficient method to reach a solution.
- **Arithmetic series formula**: Simplifies the process of adding a sequence of numbers.
- **Sum of squares formula**: Quickly calculates the sum of squared terms, useful in statistical contexts.
- **Sum of cubes formula**: Provides an efficient way to determine the sum of cube values.
They are integral to disciplines such as engineering, economics, computer science, and more, demonstrating the universal language of mathematics.