Chapter 14: Problem 25
Are the expressions in Problems equivalent? \(\sum_{i=1}^{20}(-1)^{i} b_{n}\) and \(-b_{1}+b_{2}+b_{3}+\cdots+b_{19}+b_{20}\)
Short Answer
Expert verified
Answer: Yes, the two expressions are equivalent.
Step by step solution
01
Understand the first expression
The first expression \(\sum_{i=1}^{20}(-1)^{i} b_{n}\) represents a sum of 20 terms, where each term is of the form \((-1)^{i} b_{n}\). Notice that the exponent of \((-1)\) is equal to the index \(i\) in the sum, which means that we alternate between positive and negative terms.
02
Expand the first expression
Let's expand the first expression to understand its terms better.
\(\sum_{i=1}^{20}(-1)^{i} b_{n} = (-1)^1 b_1 + (-1)^2 b_2 + (-1)^3 b_3 + \cdots + (-1)^{19} b_{19} + (-1)^{20} b_{20}\).
03
Simplify the first expression
We can now simplify this expression by recognizing that \((-1)^{odd}\) will always be \(-1\) and \((-1)^{even}\) will always be \(1\). The expression now becomes:
\(-b_1 + b_2 - b_3 + \cdots - b_{19} + b_{20}\).
04
Compare the second expression
The second expression is given as \(-b_{1}+b_{2}+b_{3}+\cdots+b_{19}+b_{20}\). We can notice that the terms in both the expressions match.
Based on the above steps, we can conclude that the two given expressions, \(\sum_{i=1}^{20}(-1)^{i} b_{n}\) and \(-b_{1}+b_{2}+b_{3}+\cdots+b_{19}+b_{20}\), are equivalent.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sigma Notation
Sigma notation, denoted as \( \sum \), is a concise way of writing sums, particularly those with many terms. It represents the summation of a sequence of numbers or expressions. When you see sigma notation, the expression underneath, above, and beside the sigma will guide you on what to sum.
In our problem, sigma notation helps easily express the process of alternating the terms as we switch each factor from negative to positive based on the exponent \( i \), which simplifies complex expressions and aids comprehension.
- Below the sigma is the starting index, in this case, \( i=1 \).
- Above the sigma is the stopping index, which is \( 20 \) in the exercise.
- Beside the sigma is the expression to be summed, here, \((-1)^{i} b_{n}\).
In our problem, sigma notation helps easily express the process of alternating the terms as we switch each factor from negative to positive based on the exponent \( i \), which simplifies complex expressions and aids comprehension.
Alternating Series
An alternating series is a sequence of numbers where the terms are alternately positive and negative. It's easy to identify because the signs "alternate," or switch back and forth.
Alternating series are common in mathematical analysis and play a significant role in calculus, especially in series convergence and approximation of functions. Being able to identify the alternating pattern is key to understanding and simplifying expressions.
- For the given exercise, \((-1)^{i}\) determines the sign of each term. If \( i \) is odd, \((-1)^{i} = -1\), making the term negative.
- If \( i \) is even, \((-1)^{i} = 1\), making the term positive.
Alternating series are common in mathematical analysis and play a significant role in calculus, especially in series convergence and approximation of functions. Being able to identify the alternating pattern is key to understanding and simplifying expressions.
Simplification of Expressions
Simplifying expressions involves rewriting them in a way that is easily understandable and more concise without changing their value.
In mathematical problems, especially with sigma notation exercises, simplification can reveal equivalent expressions that look different at first glance.
In mathematical problems, especially with sigma notation exercises, simplification can reveal equivalent expressions that look different at first glance.
- The original expression \( \sum_{i=1}^{20} (-1)^{i} b_{n} \) simplifies by recognizing that \((-1)\) raised to odd and even powers will alternate the terms' signs.
- Through simplification, it transforms into \(-b_1 + b_2 - b_3 + \ldots - b_{19} + b_{20}\).