Problem 19 - Thomas Calculus

Question

Evaluate the sums in Exercises \(19-28\). $$ \begin{array}{lll}{\text { a. } \sum_{k=1}^{10} k} & {\text { b. } \sum_{k=1}^{10} k^{2}} & {\text { c. } \sum_{k=1}^{10} k^{3}}\end{array} $$

Step-by-Step Solution

Answer

a. 55, b. 385, c. 3025

Step 1: Evaluate \(\sum_{k=1}^{10} k\)

To find the sum \(\sum_{k=1}^{10} k\), use the formula for the sum of the first \(n\) natural numbers: \(\frac{n(n+1)}{2}\). Here, \(n = 10\), so the sum is \(\frac{10 \cdot (10+1)}{2} = \frac{10 \cdot 11}{2} = 55\).

Step 2: Evaluate \(\sum_{k=1}^{10} k^2\)

To evaluate \(\sum_{k=1}^{10} k^2\), use the formula for the sum of squares of the first \(n\) natural numbers: \(\frac{n(n+1)(2n+1)}{6}\). Substituting \(n = 10\), the sum is \(\frac{10 \cdot 11 \cdot 21}{6} = 385\).

Step 3: Evaluate \(\sum_{k=1}^{10} k^3\)

The sum \(\sum_{k=1}^{10} k^3\) can be computed using the formula: \(\left(\frac{n(n+1)}{2}\right)^2\). Again, \(n = 10\), so the sum is \(\left(\frac{10 \cdot 11}{2}\right)^2 = 55^2 = 3025\).

Key Concepts

Sum of Natural Numbers

The sum of natural numbers represents the total when you add together a sequence of natural numbers starting from 1 up to a certain number, which we call \( n \). This can be easily calculated using the formula for the sum of the first \( n \) natural numbers:

• Formula: \( \frac{n(n+1)}{2} \)

To see how this works, consider the case where \( n = 10 \). By using the formula, you substitute 10 for \( n \), leading to \( \frac{10 imes 11}{2} \). Simplifying this gives you 55.
It’s fascinating because the formula is derived from pairing numbers equidistant from the beginning and the end of the sequence, like \( 1 + 10, 2 + 9, 3 + 8, \) and so on, which are all equal to 11. Thus for 10 terms, there are 5 such pairs.

Sum of Squares

The sum of squares involves adding up the squares of each number in the sequence from 1 to \( n \). Squaring a number means multiplying the number by itself. The formula used to calculate the sum of squares for the first \( n \) natural numbers is:

• Formula: \( \frac{n(n+1)(2n+1)}{6} \)

For \( n = 10 \), substituting these values into the formula gives \( \frac{10 imes 11 imes 21}{6} \). Working through this multiplication and division results in 385.
The sum of squares grows quite quickly because each number is squared before adding, illustrating the increasing difference in value as numbers get larger. This formula is fantastic for quickly finding sums from larger series without having to manually square and add each one.

Sum of Cubes

The sum of cubes is the total you get when each number from 1 to \( n \) is cubed, or raised to the power of three, and then added together. The formula for calculating the sum of cubes is unique because it relates directly to the sum of natural numbers:

• Formula: \( \left(\frac{n(n+1)}{2}\right)^2 \)

With \( n = 10 \), you first find the sum of natural numbers (which is 55, as calculated previously), and then square this result: \( 55^2 = 3025 \).
What’s interesting here is the sum of cubes formula being a perfect square. This ties together elegantly with the arithmetic series since each cubed sum is essentially expanding a natural sum square. Such beautiful patterns are common in mathematics, especially when exploring series and sequences.