Problem 94 - Elementary and Intermediate Algebra

Question

Evaluate each expression. See Example 10. $$ \frac{a\left(1-r^{n}\right)}{1-r} \text { for } a=-5, r=2, \text { and } n=3 $$

Step-by-Step Solution

Answer

The expression evaluates to -35.

Step 1: Understand the formula

We are given the formula for the sum of a geometric series: \( \frac{a(1-r^n)}{1-r} \). This formula is used when \( r eq 1 \). Here, \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.

Step 2: Substitute the values

Substitute the given values into the formula: \( a = -5 \), \( r = 2 \), and \( n = 3 \). This gives us: \[ \frac{-5 (1-2^3)}{1-2} \].

Step 3: Simplify inside the brackets

Calculate \( 2^3 = 8 \), so the expression inside the brackets becomes \( 1 - 8 = -7 \). We now have: \[ \frac{-5 (-7)}{1-2} \].

Step 4: Simplify the denominator

The denominator \( 1 - 2 = -1 \). So the expression simplifies to \[ \frac{-5 (-7)}{-1} \].

Step 5: Calculate the numerator

Multiply \(-5\) by \(-7\) to get \(35\). Thus, we have: \[ \frac{35}{-1} \].

Step 6: Divide to get the final result

Divide \(35\) by \(-1\) to get \(-35\). Hence, the value of the expression is \(-35\).

Key Concepts

Evaluation of Expressions

When evaluating mathematical expressions, especially those involving a formula, the key is to substitute the given values into the equation. This helps simplify the equation and solve it directly. For example, if you have the expression \( \frac{a(1-r^n)}{1-r} \), and values are provided for \( a \), \( r \), and \( n \), your first task is to replace these variables with the provided numbers. This substitution transforms the expression into a specific numerical problem.

• Start by identifying each variable from the expression.
• Replace each variable with the corresponding number.
• Use the order of operations (parentheses, exponents, multiplication and division, addition and subtraction) to evaluate the entire expression.

Through this substitution and following the correct mathematical operations, you can simplify complex algebraic expressions into a single numerical answer. This process is especially useful in solving geometric series problems.

Substitution Methods

Substitution methods are a fundamental tool in algebra that allow us to replace variables with their actual values. This is crucial for evaluating expressions effectively, especially in problems involving sequences or series like the geometric series.

• Identify the expression and understand the role of each variable.
• Carefully substitute each variable with the given number.
• Make sure to respect the position and operation associated with each variable during substitution.

For the expression \( \frac{a(1-r^n)}{1-r} \), substituting the given values \( a = -5, \; r = 2, \; n = 3 \) transforms it into a completely numerical problem \( \frac{-5(1-8)}{1-2} \). Ensure you work carefully through each step, double-checking your substitutions, to avoid common errors.Substitution assists in converting algebraic expressions into straightforward arithmetic operations, making complex problems much simpler to solve.

Algebraic Simplification

Algebraic simplification is vital for making expressions readable and solvable. It involves reducing an expression to its simplest form. After substitution in a problem like the geometric series, simplification helps break down the complexities.

• First, simplify any expressions inside parentheses or brackets.
• Calculate powers or exponents next, as shown in our example with \( 2^3 = 8 \).
• Continue simplifying by performing multiplication, division, addition, or subtraction in the required order.

For the expression \( \frac{-5(-7)}{-1} \):

1. Simplify the parentheses: \(1 - 8 = -7\).
2. Multiply: \(-5 \times -7 = 35\).
3. Divide: \(\frac{35}{-1} = -35\).

Every simplification step brings you closer to the final result, ensuring accuracy in calculations. Proper simplification is the backbone of solving algebraic expressions, leading to answers like \(-35\) in our original example.