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Chapter 12: Problem 37

Find the first four partial sums and the \(n\)th partial sum of the sequence \(a_{n} .\) \(a_{n}=\sqrt{n}-\sqrt{n+1}\)

Short Answer

Expert verified
The nth partial sum is \(S_n = \sqrt{1} - \sqrt{n+1}\).

Step by step solution

01

Find the First Partial Sum

The first partial sum, denoted as \(S_1\), is simply the first term of the sequence \(a_n\). Calculate \(S_1 = a_1 = \sqrt{1} - \sqrt{2}\).
02

Find the Second Partial Sum

The second partial sum, \(S_2\), is the sum of the first two terms: \(S_2 = a_1 + a_2\). To compute \(a_2\), substitute \(n = 2\) into the sequence: \(a_2 = \sqrt{2} - \sqrt{3}\). Now, calculate \(S_2 = (\sqrt{1} - \sqrt{2}) + (\sqrt{2} - \sqrt{3})\). The middle terms cancel each other, so \(S_2 = \sqrt{1} - \sqrt{3}\).
03

Find the Third Partial Sum

The third partial sum, \(S_3\), is the sum of the first three terms: \(S_3 = a_1 + a_2 + a_3\). Calculate \(a_3\) using \(n = 3\): \(a_3 = \sqrt{3} - \sqrt{4}\). Thus, \(S_3 = (\sqrt{1} - \sqrt{2}) + (\sqrt{2} - \sqrt{3}) + (\sqrt{3} - \sqrt{4})\). The terms cancel out to \(S_3 = \sqrt{1} - \sqrt{4}\).
04

Find the Fourth Partial Sum

The fourth partial sum, \(S_4\), adds the first four terms: \(S_4 = a_1 + a_2 + a_3 + a_4\). Use \(n = 4\) to find \(a_4 = \sqrt{4} - \sqrt{5}\). So, \(S_4 = (\sqrt{1} - \sqrt{2}) + (\sqrt{2} - \sqrt{3}) + (\sqrt{3} - \sqrt{4}) + (\sqrt{4} - \sqrt{5})\). The terms simplify to \(S_4 = \sqrt{1} - \sqrt{5}\).
05

Generalize to the nth Partial Sum

The \(n\)th partial sum, denoted as \(S_n\), is the sum of the first \(n\) terms: \(S_n = a_1 + a_2 + \ldots + a_n\). Observe that each term \(a_i\) cancels out except for the first and last term: \(S_n = \sqrt{1} - \sqrt{n+1}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Sequences
A sequence in mathematics is simply an ordered list of numbers. Imagine you have a series of steps leading to a destination, each number in a sequence is a step in this ordered path. Sequences can be finite or infinite and are often expressed with a general formula. This formula allows you to determine any term in the sequence. An example formula is the one from our exercise, where the term is defined as \( a_n = \sqrt{n} - \sqrt{n+1} \).
  • A sequence could be as simple as 1, 2, 3, 4, which is called an arithmetic sequence with a common difference of 1.
  • Another common type is a geometric sequence, like 2, 4, 8, 16, which multiplies each term by a common ratio.
Each specific number in the sequence is called a term, and the position of that term is indicated by the subscript \(n\). Understanding sequences is fundamental to grasping more complex mathematical concepts such as series and sums.
Exploring Series
When we talk about a series, we are looking at the process of adding up the terms of a sequence. Imagine adding up all the steps you've taken鈥攖he sum total is what we call a series.
  • A series can be finite, where you calculate the sum of a specific number of terms.
  • It can also be infinite, continuing forever and represented as a limit to find its total.
In our example, when we add the first four terms of \( a_n = \sqrt{n} - \sqrt{n+1} \), we form a partial sum, which is a specific type of series. Each partial sum builds upon the previous, showing the gradual accumulation of the sequence's terms. The goal in many problems is to find the general sum of \( n \) terms, which requires recognizing patterns or simplifications like the telescoping we see here.
Understanding Telescoping Series
In the example of our exercise with the sequence \( a_n = \sqrt{n} - \sqrt{n+1} \), we witness a fascinating concept called a telescoping series. Telescoping series are special because many of their internal terms cancel out.
Consider how the series is structured:
  • When you add successive terms, much of the sequence 'folds' or 'collapses' into a simpler expression.
  • In our case, each \( \sqrt{n} \) is canceled by the following \( -\sqrt{n} \) in the next term, leaving only the initial \( \sqrt{1} \) and the final \( -\sqrt{n + 1} \).
This clever cancelation means that instead of summing numerous terms, you end up with a very manageable expression, such as \( S_n = \sqrt{1} - \sqrt{n+1} \). Telescoping is not only elegant but extremely useful in mathematical simplification.
Applying Concepts in Mathematics Education
Mathematics education aims to take abstract concepts like sequences and series and make them accessible to everyone. The step-by-step practice as shown in our exercise is one such approach.
  • Students start with concrete examples, slowly piecing together the logic behind the math.
  • Telescoping series and partial sums also develop problem-solving skills鈥攁llowing for recognizing patterns and applying strategic simplifications.
Education emphasizes understanding over mere memorization. This approach ensures that learners can apply knowledge to new problems, growing beyond the classroom. Solving exercises like these gives students insight into how math works in layers, each one building on the last, to form a complete understanding.

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Most popular questions from this chapter

\(13-20=\) Evaluate the expression. $$ \left(\begin{array}{l}{5} \\ {0}\end{array}\right)+\left(\begin{array}{l}{5} \\\ {1}\end{array}\right)+\left(\begin{array}{l}{5} \\\ {2}\end{array}\right)+\left(\begin{array}{l}{5} \\\ {3}\end{array}\right)+\left(\begin{array}{l}{5} \\\ {4}\end{array}\right)+\left(\begin{array}{l}{5} \\ {5}\end{array}\right) $$

Find the term containing \(y^{3}\) in the expansion of \((\sqrt{2}+y)^{12}\)

Express the repeating decimal as a fraction. $$ 2.11 \overline{25} $$

\(21-24\) Use the Binomial Theorem to expand the expression. $$ \left(2 A+B^{2}\right)^{4} $$

\(1-12\) . Use Pascal's triangle to expand the expression. $$ (2 x+1)^{4} $$
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