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Chapter 3: Problem 10

Obtain a formula for $$ 2+2 \sqrt{2}+4+4 \sqrt{2}+8+8 \sqrt{2}+\cdots+2^{m} . $$

Short Answer

Expert verified
The sum is \( S = 2^m + 2^m\sqrt{2} - 1 - 2\sqrt{2} \).

Step by step solution

01

Identify Pattern in Terms

First, observe the expression: \[ 2 + 2\sqrt{2} + 4 + 4\sqrt{2} + 8 + 8\sqrt{2} + \cdots + 2^m. \]We can see that this is a combination of powers of 2 and \( \sqrt{2} \):- Terms without \( \sqrt{2}\) are powers of 2: \( 2, 4, 8, \ldots, 2^{m}. \)- Terms with \( \sqrt{2}\) are twice the previous power, e.g., \( 2\sqrt{2}, 4\sqrt{2}, 8\sqrt{2}, \ldots, 2^{m-1}\sqrt{2}. \)This sequence can be split into two separate geometric series.
02

Express Sequences as Geometric Series

Write each sub-sequence as a separate geometric series:1. Sequence without \( \sqrt{2} \): \[ 2 + 4 + 8 + \cdots + 2^m. \] This is a geometric series with first term \( a = 2 \) and common ratio \( r=2 \).2. Sequence with \( \sqrt{2} \): \[ 2\sqrt{2} + 4\sqrt{2} + 8\sqrt{2} + \cdots + 2^{m-1}\sqrt{2}. \] This is another geometric series with first term \( a' = 2\sqrt{2} \) and common ratio \( r=2 \).
03

Use the Sum Formula for Geometric Series

The sum of a geometric sequence is given by:\[ S_n = a \frac{r^n - 1}{r - 1}, \]where \( n \) is the number of terms, \( a \) is the first term, and \( r \) is the common ratio.For the first sequence:- First term \( a = 2 \), ratio \( r = 2 \), \( n = m. \)- Sum is \[ S_1 = 2 \frac{2^m - 1}{2 - 1} = 2^m - 1. \]For the second sequence:- First term \( a' = 2\sqrt{2} \), ratio \( r = 2 \), and \( n = m. \)- Sum is \[ S_2 = 2\sqrt{2} \frac{(2^{m - 1}) - 1}{2 - 1} = 2^{m} \sqrt{2} - 2\sqrt{2}. \]
04

Combine Series Results

Add both sums to find the total sum:\[ S = (2^m - 1) + (2^m\sqrt{2} - 2\sqrt{2}). \]Simplify this expression:\[ S = 2^m + 2^m\sqrt{2} - 1 - 2\sqrt{2}. \]Therefore, the total expression for the sum of the series is:\[ S = 2^m + 2^m\sqrt{2} - 1 - 2\sqrt{2}. \]

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Key Concepts

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

Sequence
A sequence is basically a list of numbers arranged in a particular order. Sequences come in various types, and one common type is a geometric sequence, where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the "common ratio".
This exercise is centered around creating and solving a sequence that combines terms with and without the square root of 2. Let's break it down:
  • The sequence without the square root of 2 is straightforward and looks like this: 2, 4, 8, ..., up to the last term \( 2^m \). Every term is a power of 2.
  • The sequence with the square root of 2 is similar, but each term is multiplied by \( \sqrt{2} \). It goes: \( 2\sqrt{2}, 4\sqrt{2}, 8\sqrt{2}, \ldots, 2^{m-1}\sqrt{2} \).
This way, you can visualize how two separate sequences are identified and can be treated individually in calculations.
Powers of Two
The term 'powers of two' refers to numbers that can be expressed in the form \( 2^n \), where \( n \) is a non-negative integer. In this context, the exercise deals with sequences made up of such numbers.
Understanding powers of two is essential because it shows how rapidly numbers grow:
  • \( 2^0 \) is 1
  • \( 2^1 \) is 2
  • \( 2^2 \) is 4
  • \( 2^3 \) is 8, and so forth
In the problem at hand, powers of two appear in both the main sequence and the sequence involving the square root of 2. By multiplying these powers of two by \( \sqrt{2} \), we alter, yet maintain, the structured progression of these terms within the full series.
Sum Formula
To find the total sum of a geometric sequence, the sum formula for geometric series comes into play. This formula helps calculate the sum efficiently without adding each term manually. The formula is given by:\[ S_n = a \frac{r^n - 1}{r - 1} \]where \( S_n \) is the sum of the first \( n \) terms, \( a \) is the first term of the sequence, and \( r \) is the common ratio.
The exercise provided two series to combine:
  • First, the series without the square root \( 2 \) uses this sum formula to give \( 2^m - 1 \).
  • Second, the series with \( \sqrt{2} \) involves a multiplication within the formula, leading to the sum \( 2^{m}\sqrt{2} - 2\sqrt{2} \).
By using these calculated sums, they bring together both series into one cohesive result, showing how all terms interconnect within the original expression.

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Most popular questions from this chapter

Let \(\left\\{a_{k}\right\\}\) be a monotonic sequence of real numbers such that \(\sum_{k=1}^{\infty} a_{k}\) converges. Show that $$ \sum_{k=1}^{\infty} k\left(a_{k}-a_{k+1}\right) $$ converges.

For any two series of positive terms write $$ \sum_{k=1}^{\infty} a_{k} \preceq \sum_{k=1}^{\infty} b_{k} $$ if \(a_{k} / b_{k} \rightarrow 0\) as \(k \rightarrow \infty\) (a) If both series converge, explain why this might be interpreted by saying that \(\sum_{k=1}^{\infty} a_{k}\) is converging faster than \(\sum_{k=1}^{\infty} b_{k}\). (b) If both series diverge, explain why this might be interpreted by saying that \(\sum_{k=1}^{\infty} a_{k}\) is diverging more slowly than \(\sum_{k=1}^{\infty} b_{k}\). (c) For convergent series is there any connection between $$ \sum_{k=1}^{\infty} a_{k} \preceq \sum_{k=1}^{\infty} b_{k} $$ and $$ \sum_{k=1}^{\infty} a_{k} \leq \sum_{k=1}^{\infty} b_{k} ? $$ (d) For what values of \(p, q\) is $$ \sum_{k=1}^{\infty} \frac{1}{k^{p}} \preceq \sum_{k=1}^{\infty} \frac{1}{k^{q}} ? $$ (e) For what values of \(r, s\) is $$ \sum_{k=1}^{\infty} r^{k} \preceq \sum_{k=1}^{\infty} s^{k} ? $$ (f) Arrange the divergent series $$ \sum_{k=2}^{\infty} \frac{1}{k}, \sum_{k=2}^{\infty} \frac{1}{k \log k}, \sum_{k=2}^{\infty} \frac{1}{k \log (\log k)}, \sum_{k=2}^{\infty} \frac{1}{k \log (\log (\log k))} \ldots $$ into the correct order. (g) Arrange the convergent series $$ \begin{gathered} \sum_{k=2}^{\infty} \frac{1}{k^{p}}, \sum_{k=2}^{\infty} \frac{1}{k(\log k)^{p}}, \sum_{k=2}^{\infty} \frac{1}{k \log k(\log (\log k))^{p}} \\ \sum_{k=2}^{\infty} \frac{1}{k \log k(\log (\log k))(\log (\log (\log k)))^{p}} \ldots \end{gathered} $$ into the correct order. Here \(p>1\). (h) Suppose that \(\sum_{k=1}^{\infty} b_{k}\) is a divergent series of positive numbers. Show that there is a series $$ \sum_{k=1}^{\infty} a_{k} \preceq \sum_{k=1}^{\infty} b_{k} $$ that also diverges (but more slowly). (i) Suppose that \(\sum_{k=1}^{\infty} a_{k}\) is a convergent series of positive numbers. Show that there is a series $$ \sum_{k=1}^{\infty} a_{k} \preceq \sum_{k=1}^{\infty} b_{k} $$ that also converges (but more slowly). (j) How would you answer this question? Is there a "mother" of all divergent series diverging so slowly that all other divergent series can be proved to be divergent by a comparison test with that series?

If $$ s_{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4} \cdots+\cdots+\frac{1}{n} $$ show that \(s_{2^{n}} \geq 1+n / 2\) for all \(n\).

The sum rule for convergent series $$ \sum_{k=1}^{\infty}\left(a_{k}+b_{k}\right)=\sum_{k=1}^{\infty} a_{k}+\sum_{k=1}^{\infty} b_{k} $$ can be expressed by saying that if any two of these series converges so too does the third. What kind of statements can you make for absolute convergence and for nonabsolute convergence?

Let \(\left\\{s_{n}\right\\}\) be any sequence of real numbers. Show that this sequence converges to a number \(S\) if and only if the series $$ s_{1}+\sum_{k=2}^{\infty}\left(s_{n}-s_{n-1}\right) $$ converges and has sum \(S\).
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