Problem 37 - Algebra and Trigonometry with Analytic Geometry

Question

\(\sum_{k=0}^{5} k(k-2) \quad\)

Step-by-Step Solution

Answer

The result of the summation is 25.

Step 1: Expand the expression

First, expand the expression inside the summation. Given the expression is \( k(k-2) \), expand it to \( k^2 - 2k \). This will allow us to separate the terms when finding the summation.

Step 2: Distribute the summation

Apply the summation to each part of the expanded expression. So, the original summation \( \sum_{k=0}^{5} k(k-2) \) becomes two separate summations: \( \sum_{k=0}^{5} k^2 - \sum_{k=0}^{5} 2k \). This is allowed because of the linearity property of summation.

Step 3: Calculate each summation

Calculate \( \sum_{k=0}^{5} k^2 \) and \( \sum_{k=0}^{5} k \) separately. Use known formulas: - \( \sum_{k=0}^{n} k = \frac{n(n+1)}{2} \) - \( \sum_{k=0}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \).Thus, substituting \( n = 5 \) gives us:\( \sum_{k=0}^{5} k = \frac{5 \times 6}{2} = 15 \) \( \sum_{k=0}^{5} k^2 = \frac{5 \times 6 \times 11}{6} = 55 \).

Step 4: Substitute and simplify

Now substitute the results back into the distributed summation: \( \sum_{k=0}^{5} (k^2 - 2k) \) equals \( \sum_{k=0}^{5} k^2 - 2 \sum_{k=0}^{5} k \). Substitute to get \( 55 - 2 \times 15 \). Simplify this to obtain the final result \( 55 - 30 = 25 \).

Key Concepts

Linearity Property of Summation

The linearity property of summation is a very useful technique when working with sums or sequences. It allows us to simplify complex summations by splitting them into smaller, more manageable parts. This means if you have a sum of a combination of terms, like the expression

• \( \sum (a_k + b_k) = \sum a_k + \sum b_k \),
• you can separate each term.

This property doesn't stop at addition, it works with subtraction too! For instance, in our exercise, the summation \( \sum_{k=0}^{5} k^2 - 2k \) was divided into two parts: the sum of squares \( \sum_{k=0}^{5} k^2 \) and the sum of integers \( 2\sum_{k=0}^{5} k \). This approach is crucial because each separated part can be tackled with different formulas or methods, leading to a much simpler solution process. So, remember, breaking it down is key!

Sum of Squares Formula

The sum of squares formula is a special formula used to find the sum of the squares of a sequence of numbers. In mathematics, the formula for the sum of squares of the first \( n \) integers is expressed as:

• \( \sum_{k=0}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \).

This formula is particularly useful when dealing with quadratic terms like \( k^2 \). In our original problem, we calculated the sum of squares from 0 to 5. By substituting \( n = 5 \) into the formula, we efficiently found that:

• \( \sum_{k=0}^{5} k^2 = \frac{5 \times 6 \times 11}{6} = 55 \).

Notice how quickly the result is derived using the formula, saving a lot of time compared to adding manually! Using these results from formulas allows us to focus on the big picture and understand the structure of mathematical problems.

Sum of Integers Formula

Calculating the sum of integers is straightforward if you use the sum of integers formula, which is:

• \( \sum_{k=0}^{n} k = \frac{n(n+1)}{2} \).

This formula efficiently gives us the total of all integers from 1 through \( n \). In this exercise's solution, it was applied to find the sum of the integers from 0 to 5. After substituting \( n = 5 \), we found that:

• \( \sum_{k=0}^{5} k = \frac{5 \times 6}{2} = 15 \).

The simplicity of the formula helps cultivate a better understanding of linear sums and their relationships. Like slicing a pie, each sum is a piece of the puzzle that comes together to solve the bigger problem. So, next time you encounter a sequence of integers, remember that the sum of integers formula is a go-to tool for quick and easy calculations.