Problem 26 - Thomas Calculus

Question

Evaluate the sums in Exercises \(19-28\). $$ \sum_{k=1}^{7} k(2 k+1) $$

Step-by-Step Solution

Answer

The sum evaluates to 308.

Step 1: Expand the Expression

The given sum is \( \sum_{k=1}^{7} k(2k+1) \). First, we expand the expression inside the summation: \[ k(2k+1) = 2k^2 + k \]. Now, the sum becomes: \[ \sum_{k=1}^{7} (2k^2 + k) \].

Step 2: Split the Summation

The sum \( \sum_{k=1}^{7} (2k^2 + k) \) can be split into two separate sums:\[ \sum_{k=1}^{7} 2k^2 + \sum_{k=1}^{7} k \]. Let's proceed with these two separate sums.

Step 3: Evaluate the First Sum

For \( \sum_{k=1}^{7} 2k^2 \), we can factor out the 2:\[ 2 \sum_{k=1}^{7} k^2 \]. The formula for the sum of squares is \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \). Substitute \( n = 7 \):\[ \sum_{k=1}^{7} k^2 = \frac{7 \times 8 \times 15}{6} = 140 \]. Thus, \( 2 \sum_{k=1}^{7} k^2 = 2 \times 140 = 280 \).

Step 4: Evaluate the Second Sum

Now evaluate \( \sum_{k=1}^{7} k \), which has the formula \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \). Substitute \( n = 7 \):\[ \sum_{k=1}^{7} k = \frac{7 \times 8}{2} = 28 \].

Step 5: Combine the Results

We now combine the results of the two sums:\[ 280 + 28 = 308 \]. Therefore, the value of the original sum \( \sum_{k=1}^{7} k(2k+1) \) is 308.

Key Concepts

Sum of Squares Formula

When dealing with series summation, especially those involving squares of numbers, the sum of squares formula becomes quite essential. This formula allows us to find the sum of squares of a sequence of integers efficiently. The formula for obtaining the sum of the squares of the first \( n \) natural numbers is:
\[\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\]Using this formula, we can quickly calculate the sum without having to square each number sequentially and then add them up. For example, for summing the squares from 1 to 7, plug \( n = 7 \) into the formula:

• Multiply \( n \), \( n+1 \), and \( 2n+1 \) together: \( 7 \times 8 \times 15 \).
• Divide the product by 6 to get 140.

The sum of squares formula saves time and ensures accuracy when dealing with larger series.

Arithmetic Series

An arithmetic series, also known as an arithmetic progression, involves a sequence of numbers with a common difference between consecutive terms. To calculate the sum of such a series up to \( n \) terms, we use a straightforward formula:
\[\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\]This formula is very handy for quickly summing a linear series of numbers. In our exercise, using \( n = 7 \), the sum of the first 7 natural numbers is calculated as:

• Multiply \( n = 7 \) by \( n+1 = 8 \) to get 56.
• Divide by 2 to obtain 28.

The sum of these numbers results in a quick and easy total that is frequently used as part of larger algebraic calculations.

Algebraic Manipulation

Algebraic manipulation is an essential skill in mathematics, allowing us to reshape and simplify expressions or solve equations more easily. In series summation, the ability to manipulate algebraic expressions helps break down complex problems into manageable parts.
Consider the original expression in our exercise, \( \sum_{k=1}^{7} k(2k+1) \). By distributing the \( k \) through the parentheses, we convert it into \( 2k^2 + k \), simplifying our approach. This step then allows us to split the summation into two separate sum components: \( \sum_{k=1}^{7} 2k^2 \) and \( \sum_{k=1}^{7} k \).

• Transformation using distribution helps simplify complex expressions.
• Allows separate evaluation using known formulas for sums of series.

With practice, algebraic manipulation becomes a powerful tool for dealing with various mathematical problems.

Mathematical Induction

Mathematical induction is a method used to prove hypotheses or theories that are formulated in terms of natural numbers. It is especially useful in proving the validity of formulas used in series summation, like the sum of squares or arithmetic formulas.
The process generally consists of two steps:

• Base Case: Verify that the formula holds for the initial value, often \( n = 1 \).
• Inductive Step: Assume the formula is valid for some arbitrary natural number \( n \), and then prove it for \( n+1 \).

By completing these steps, we can affirm the truth of the mathematical formula for all natural numbers, giving us the confidence to use these formulas in exercises like the one we’ve seen. Although not used directly in the problem, understanding induction helps appreciate the power and reliability of these formulas.