Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use Special Sum Formulas 1-4 to find each sum. $$ \sum_{i=1}^{10}[(i-1)(4 i+3)] $$

Step-by-Step Solution

Answer

The sum is 1455.

Step 1: Expand the Expression Within the Summation

First, we expand the expression \[(i-1)(4i + 3)\].Distribute each term in \((i-1)\) to the terms in \((4i + 3)\):\[= i(4i) + i(3) - 1(4i) - 1(3)\]\[= 4i^2 + 3i - 4i - 3\]Simplify:\[= 4i^2 - i - 3\].Thus, \[\sum_{i=1}^{10}[(i-1)(4i+3)] = \sum_{i=1}^{10}[4i^2 - i - 3].\]

Step 2: Separate the Summation

The next step is to separate the single summation into three separate summations:\[\sum_{i=1}^{10}[4i^2 - i - 3] = 4\sum_{i=1}^{10}i^2 - \sum_{i=1}^{10}i - \sum_{i=1}^{10}3.\]Each of these summations can be calculated using known formulas.

Step 3: Apply the Special Sum Formulas

Now, apply the special sum formulas:For\(\sum_{i=1}^{n} i^2\), the formula is: \[\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\].Substitute \(n = 10\):\[\sum_{i=1}^{10} i^2 = \frac{10(10+1)(2 \times 10+1)}{6} = \frac{10 \times 11 \times 21}{6} = 385.\]

Step 4: Calculate Linear and Constant Sums

The sum of integers from 1 to 10 is given by:For\(\sum_{i=1}^{n} i\), the formula is: \[\sum_{i=1}^{n} i = \frac{n(n+1)}{2}\].Substitute \(n = 10\):\[\sum_{i=1}^{10} i = \frac{10 \times 11}{2} = 55.\]The sum of the constant \(3\) for 10 terms is:\[\sum_{i=1}^{10} 3 = 3 \times 10 = 30.\]

Step 5: Substitute the Calculated Sums Back

Substitute the calculated sums back into the separated equation:\[4\sum_{i=1}^{10}i^2 - \sum_{i=1}^{10}i - \sum_{i=1}^{10}3\]\[= 4 \times 385 - 55 - 30\].Calculate each term: \[= 1540 - 55 - 30 = 1455.\]

Step 6: Final Result

The summation results in\[1455\].Thus,\[\sum_{i=1}^{10}[(i-1)(4i+3)] = 1455.\]

Key Concepts

Special Sum Formulas

Special sum formulas are mathematical shortcuts used to calculate sums of numeric sequences quickly. Rather than adding each individual number, these formulas utilize patterns in sequences to provide an efficient solution. For example:

• The sum of the first \(n\) positive integers is calculated using \( \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \).
• The sum formula for squares, \( \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \), helps tackle quadratic terms.
• For constant terms like a repetitive number \(a\), the sum is just \( \sum_{i=1}^{n} a = a \times n \).

These formulas save time and reduce calculation errors. Understanding the repeatable nature of these sums can greatly assist with larger or more complex problems.

Polynomial Expansion

The technique of polynomial expansion involves breaking down products of expressions into simpler terms using algebraic rules. For expanding the expression \((i-1)(4i+3)\), we rely on distributing the multiplicands across all terms:

• Multiply each term in \((i-1)\) with each term in \((4i+3)\).
• Combine similar terms to reveal the expanded form: \(i \cdot 4i + i \cdot 3 - 1 \cdot 4i - 1 \cdot 3 = 4i^2 + 3i - 4i - 3 = 4i^2 - i - 3\).

Expanding polynomials is essential to simplifying and solving more complex algebraic expressions and assists in accurately setting up summation problems.

Distributive Property

The distributive property is a foundational principle in algebra, stating that multiplication distributes over addition. This is expressed as \( a(b + c) = ab + ac \). In practice, the distributive property helps simplify expressions, like expanding \((i-1)(4i+3)\):

• Multiply every term inside the parentheses by each term outside.
• This process transforms complex expressions into more manageable components.
• For instance, distributing \((i-1)\) over \((4i + 3)\) results in \(4i^2 + 3i - 4i - 3\).

This property is pivotal for simplifying algebraic expressions and constructing polynomial expansions effectively.

Integer Sequences

Integer sequences are ordered lists of integers that can form patterns useful in mathematical analysis. Sequences help us understand and calculate accumulative sums efficiently. Common types include:

• Arithmetic sequences: Sequences like 1, 2, 3, with a constant difference between terms.
• Quadratic sequences: Use special formulas when the pattern involves square numbers, such as \(i^2\).
• When summed, notably through special formulas, these sequences yield quick solutions to otherwise lengthy manual calculations.

Understanding integer sequences helps identify the type of summation or pattern involved, guiding us toward the appropriate solution method.