Problem 25 - Calculus for Biology and Medicine

Question

Use the algebraic rules for sums to evaluate each sum. Recall that $$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$$ and $$\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$$ $$ \sum_{k=0}^{6} k(k+1) $$

Step-by-Step Solution

Answer

The evaluated sum is 112.

Step 1: Understand the Sum Expression

We need to evaluate the expression \( \sum_{k=0}^{6} k(k+1) \). This is the sum of products \( k(k+1) \) from \( k=0 \) to \( k=6 \).

Step 2: Expand the Product Inside Sum

Notice that for each term, \( k(k+1) \) is the same as \( k^2 + k \). Therefore, the sum becomes \( \sum_{k=0}^{6} (k^2 + k) \).

Step 3: Split the Sum into Two Sums

Now, split the expression into two separate sums: \( \sum_{k=0}^{6} k^2 + \sum_{k=0}^{6} k \).

Step 4: Apply Known Formulas

The formulas given allow us to calculate these sums directly. However, the formulas start at \( k=1 \), not \( k=0 \). We know from the property of sums that \( \sum_{k=0}^{n} a_k = a_0 + \sum_{k=1}^{n} a_k \). This means we need to separately calculate for \( k=0 \) and then apply the formula from \( k=1 \) to \( k=6 \).

Step 5: Calculate Each Part Separately

Calculate \( \sum_{k=1}^{6} k = \frac{6(6+1)}{2} = 21 \).Calculate \( \sum_{k=1}^{6} k^2 = \frac{6(6+1)(2 \cdot 6+1)}{6} = \frac{6 \cdot 7 \cdot 13}{6} = 91 \).

Step 6: Include the Term for k=0

Now, add the terms for when \( k=0 \). Note that for \( k=0 \), both \( k^2 \) and \( k \) are 0. So, we still have 0. Thus, we only use the previously calculated values: \( 21 + 91 = 112 \).

Step 7: Final Answer

The value of the sum \( \sum_{k=0}^{6} k(k+1) \) is 112.

Key Concepts

Sum of Squares Formula

When dealing with sums that involve squares of numbers, the sum of squares formula is an essential tool. It allows us to efficiently compute the sum of squares from 1 to any positive integer, benefiting from a formula that reduces computing time.

The formula is \[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \],where the computation for every term from 1 to n is captured in this single expression.

This formula is particularly useful in mathematical series evaluations involving squares. Consider it's handy when tackling problems with quadratic expressions, like those seen in physics examinations or complex mathematical proofs.

To apply this formula:

• Identify the range of numbers, which starts at 1 and ends at your chosen n.
• Plug the value of n into the formula to get the sum quickly.

The algebraic rules for sums become much more manageable through these generalized expressions. Overall, using the sum of squares formula can simplify otherwise lengthy calculations.

Sum of Natural Numbers Formula

The sum of natural numbers formula is an algebraic shortcut to quickly add numbers in sequential order. It's a foundation of arithmetic series that helps solve many practical and theoretical problems.

This formula is expressed as:\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \].This handy formula saves time by giving the sum of the first n natural numbers without needing individual addition.

Using this formula involves a few simple steps:

• Determine your range of numbers.
• Insert the largest number (n) into the formula.
• Perform the multiplication and division to find the result.

Breaking down numbers into these succinct formulas improves understanding of larger numerical calculations, making it vital in both learning and applying mathematical concepts. In our original exercise, this formula assists with part of the solution by efficiently handling the calculations for the sum \( \sum_{k=1}^{6} k \).

Mathematical Series Evaluation

Mathematical series evaluation is about breaking down complex sums into manageable components, often using established formulas. For many mathematicians and students, evaluating a series can reveal much about number patterns and relationships.

In our context, evaluating the series of \( \sum_{k=0}^{6} k(k+1) \) involves recognizing and applying algebraic rules efficiently. Here's a step-by-step approach:

• Understand the series and the need to expand it into simpler sums.
• Decompose the expression into elementary sums, such as \( k^2 \) and \( k \).
• Use known formulas (such as the sum of squares and sum of natural numbers) to compute these individual sums.
• Add up the values obtained from these formulas to reach the total sum.

This process demonstrates the power of combining formulas to solve more complex series. It's not just about rote calculation, but understanding the interplay of individual components leading to a final, comprehensive result. By mastering these evaluations, students gain insights into the structure and behavior of series, enabling them to tackle both basic and advanced mathematical problems with confidence.