Problem 9 - Calculus Volume 1

Question

In the following exercises, use summation properties and formulas to rewrite and evaluate the sums. $$\sum_{j=1}^{50}\left(j^{2}-2 j\right)$$

Step-by-Step Solution

Answer

The sum is 40375.

Step 1: Split the Summation

The first step is to break the given summation into separate sums. We have \( \sum_{j=1}^{50} (j^2 - 2j) \). Using the sum property \( \sum (a_i - b_i) = \sum a_i - \sum b_i \), we split it into \( \sum_{j=1}^{50} j^2 - \sum_{j=1}^{50} 2j \).

Step 2: Apply Constant Multiple Rule

For the term \( \sum_{j=1}^{50} 2j \), factor out the constant 2: \( 2 \sum_{j=1}^{50} j \).

Step 3: Use Summation Formulas

Use the formulas for arithmetic series: \( \sum_{j=1}^{n} j = \frac{n(n+1)}{2} \) and \( \sum_{j=1}^{n} j^2 = \frac{n(n+1)(2n+1)}{6} \). For \( n = 50 \):1. \( \sum_{j=1}^{50} j = \frac{50 \times 51}{2} = 1275 \).2. \( \sum_{j=1}^{50} j^2 = \frac{50 \times 51 \times 101}{6} = 42925 \).

Step 4: Evaluate Individual Sums

Compute each part using the formulas:- The sum of squares: \( 42925 \).- The sum of linear terms times 2: \( 2 \times 1275 = 2550 \).

Step 5: Combine the Results

Subtract the result of the second summation from the first: \( 42925 - 2550 = 40375 \).

Key Concepts

Summation Properties

Summation properties are fundamental tools that allow us to break down complex sums into more manageable parts. When dealing with the expression \[ \sum_{j=1}^{50} (j^2 - 2j) \],we can apply the property \[ \sum (a_i - b_i) = \sum a_i - \sum b_i \].This property helps us separate the original sum into simpler sums: \[ \sum_{j=1}^{50} j^2 \] and \[ \sum_{j=1}^{50} 2j \].
In practice, these properties make computations easier and more efficient. They let us isolate different parts of an expression, apply specific formulas, and arrive at a solution systematically. Here are the key takeaways:

• You can split a sum of terms into separate summation components.
• Constant multiples can be factored out, simplifying calculations.
• These properties pave the way to apply other summation rules and formulas effectively.

Summation Formulas

Summation formulas are particularly useful when evaluating sums without calculating each term individually. They enable us to solve problems efficiently and accurately. In the context of our example, two critical summation formulas are:

• The formula for the sum of the first \( n \) integers: \[ \sum_{j=1}^{n} j = \frac{n(n+1)}{2} \]
• The formula for the sum of the squares of the first \( n \) integers: \[ \sum_{j=1}^{n} j^2 = \frac{n(n+1)(2n+1)}{6} \]

For the original problem, using these formulas provided us with quick solutions:- The sum of integers up to 50 is 1275.- The sum of squares up to 50 is 42925.
These formulas are vital for reducing workload and eliminating computational errors. By substituting directly into these formulas, you can confidently tackle large sums with ease.

Arithmetic Series

The arithmetic series formula is a pivotal component when dealing with sums involving sequences of numbers with a constant difference between consecutive terms. The formula is:\[ S_n = \frac{n}{2} \times (a + l) \]where:

• \( S_n \) is the sum of the series.
• \( n \) is the number of terms.
• \( a \) is the first term.
• \( l \) is the last term.

In our specific example involving \[ \sum_{j=1}^{50} 2j \],we recognize this as 2 times the sum of the first 50 natural numbers. Using the arithmetic series formula, we compute \( S_n \) to find \( 2 \times 1275 = 2550 \).
This shows the practical applications of the arithmetic series formula in simplifying lengthy additions into concise and straightforward calculations.

Sum of Squares

The Sum of Squares formula helps in finding the cumulative sum of squares of numbers within a defined range. It's represented as:\[ \sum_{j=1}^{n} j^2 = \frac{n(n+1)(2n+1)}{6} \]Using this formula allows you to compute intricate sums quickly. For instance, in our exercise, we used it to determine the total for squares of integers from 1 to 50, which resulted in 42925.
The significance of this formula lies in its ability to provide direct access to solutions without manual squaring and adding each term. It's crucial for problems involving quadratic patterns or when analyzing shapes and physical phenomena.

• Efficient for evaluating quadratic patterns.
• Widely applicable in physics and geometric computations.