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Magazine Chapter 13: Problem 39
\(37-40\) . Find the first four partial sums and the nth partial sum of the sequence \(a_{m}\) $$ a_{n}=\sqrt{n}-\sqrt{n+1} $$
Short Answer
Expert verified
The nth partial sum is \(S_n = \sqrt{1} - \sqrt{n+1}\).
Step by step solution
01
Calculate the First Term
For the first partial sum, substitute \(n = 1\) into the sequence formula \(a_n = \sqrt{n} - \sqrt{n+1}\). This gives the first term \(a_1 = \sqrt{1} - \sqrt{2}\).
02
Calculate the Second Partial Sum
To find the second partial sum, calculate the first two terms and add them. Using the formula, find \(a_2 = \sqrt{2} - \sqrt{3}\). The second partial sum is \(S_2 = a_1 + a_2 = (\sqrt{1} - \sqrt{2}) + (\sqrt{2} - \sqrt{3})\).
03
Calculate the Third Partial Sum
Calculate \(a_3 = \sqrt{3} - \sqrt{4}\). Add the first three terms: \(S_3 = a_1 + a_2 + a_3 = (\sqrt{1} - \sqrt{2}) + (\sqrt{2} - \sqrt{3}) + (\sqrt{3} - \sqrt{4})\).
04
Calculate the Fourth Partial Sum
Calculate \(a_4 = \sqrt{4} - \sqrt{5}\). Add together all four terms to get \(S_4 = a_1 + a_2 + a_3 + a_4 = (\sqrt{1} - \sqrt{2}) + (\sqrt{2} - \sqrt{3}) + (\sqrt{3} - \sqrt{4}) + (\sqrt{4} - \sqrt{5})\).
05
Find the General nth Partial Sum
The nth partial sum \(S_n\) is the sum of the first \(n\) terms of the sequence, which telescopes to \(S_n = \sqrt{1} - \sqrt{n+1}\). Notice how intermediate terms cancel out.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sequence
A sequence is a structured, ordered list of numbers that follow a specific rule. In the sequence we're dealing with here, the rule is defined by the formula \(a_n = \sqrt{n} - \sqrt{n+1}\). This formula tells us how to calculate each term in the sequence based on the position, \(n\), of the term.
For example:
For example:
- When \(n = 1\), the term is \(a_1 = \sqrt{1} - \sqrt{2}\).
- For \(n = 2\), the term becomes \(a_2 = \sqrt{2} - \sqrt{3}\).
- This continues indefinitely as long as you have an \(n\) value to plug into the formula.
Telescoping Series
A telescoping series is a special kind of series where many terms cancel each other out in a sequence of additions and subtractions. This makes it easier to find the sum of the series.
In our given sequence \(a_n = \sqrt{n} - \sqrt{n+1}\), when we add up the terms for partial sums, the intermediate terms cancel out. For instance:
In our given sequence \(a_n = \sqrt{n} - \sqrt{n+1}\), when we add up the terms for partial sums, the intermediate terms cancel out. For instance:
- The first two terms, \((\sqrt{1} - \sqrt{2}) + (\sqrt{2} - \sqrt{3})\), reduce to \(\sqrt{1} - \sqrt{3}\).
- Similarly, adding more terms continues this pattern, leaving only the first and the last terms.
Series Summation
Series summation involves adding the terms of a sequence together. In many mathematical problems, we are interested in figuring out the total sum of a sequence of numbers.
For the sequence \(a_n = \sqrt{n} - \sqrt{n+1}\), when we sum over the first few terms, we realize that most terms cancel out due to the telescoping nature of the series. For instance:
For the sequence \(a_n = \sqrt{n} - \sqrt{n+1}\), when we sum over the first few terms, we realize that most terms cancel out due to the telescoping nature of the series. For instance:
- To find the first four partial sums, we calculate and add each term like \(S_4 = (\sqrt{1} - \sqrt{2}) + (\sqrt{2} - \sqrt{3}) + (\sqrt{3} - \sqrt{4}) + (\sqrt{4} - \sqrt{5})\).
- After simplifying, the result dramatically simplifies due to the telescoping cancellation, giving \(\sqrt{1} - \sqrt{5}\).
nth Partial Sum
The nth partial sum is crucial when analyzing sequences and series. It represents the total sum of the first \(n\) terms of the sequence. This acts like a snapshot of the series at a certain point.
The concept is particularly significant in our exercise where the nth partial sum \(S_n = \sqrt{1} - \sqrt{n+1}\) demonstrates how intermediate terms cancel. This nth partial sum outlines the effectiveness of telescoping as:
The concept is particularly significant in our exercise where the nth partial sum \(S_n = \sqrt{1} - \sqrt{n+1}\) demonstrates how intermediate terms cancel. This nth partial sum outlines the effectiveness of telescoping as:
- The terms \(\sqrt{n} - \sqrt{n+1}\) cancel interim elements, ensuring the nth partial sum is simple and straightforward to compute.
- It provides a useful insight, showing that for any choice of \(n\), the process and calculus become straightforward.
