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Chapter 5: Problem 12

Use mathematical induction in Exercises \(3-17\) to prove summation formulae. Be sure to identify where you use the inductive hypothesis. Prove that $$ \sum_{j=0}^{n}\left(-\frac{1}{2}\right)^{j}=\frac{2^{n+1}+(-1)^{n}}{3 \cdot 2^{n}} $$ whenever \(n\) is a nonnegative integer.

Short Answer

Expert verified
Proven by induction.

Step by step solution

01

Base Case

Begin by verifying the base case where n=0. Substitute n=0 in the given formula and summation: \[ \sum_{j=0}^{0}(-\frac{1}{2})^{j} = (-\frac{1}{2})^{0} = 1 \] The right-hand side of the given formula for n=0: \[ \frac{2^{0+1}+(-1)^{0}}{3 \cdot 2^{0}} = \frac{2+1}{3 \cdot 1} = \frac{3}{3} = 1 \] Since both sides are equal, the base case holds true.
02

Inductive Step

Assume the formula is true for some arbitrary nonnegative integer k. This is the inductive hypothesis: \[ \sum_{j=0}^{k} \left(-\frac{1}{2}\right)^{j} = \frac{2^{k+1} + (-1)^{k}}{3 \cdot 2^{k}} \] Next, we need to prove it for n = k+1.
03

Inductive Proof for n = k+1

Consider \[ \sum_{j=0}^{k+1} (-\frac{1}{2})^{j} = \left( \sum_{j=0}^{k} (-\frac{1}{2})^{j} \right) + (-\frac{1}{2})^{k+1} \] Using the inductive hypothesis: \[ = \frac{2^{k+1} + (-1)^{k}}{3 \cdot 2^{k}} + (-\frac{1}{2})^{k+1} \] Simplify the right-hand side: \[ = \frac{2^{k+1} + (-1)^{k}}{3 \cdot 2^{k}} + (-\frac{1}{2})^{k+1} = \frac{2^{k+1} + (-1)^{k}}{3 \cdot 2^{k}} - \frac{1}{2^{k+1}} \] To combine the fractions, find a common denominator: \[ = \frac{2^{k+1} + (-1)^{k} - 3}{3 \cdot 2^{k}} = \frac{2^{k+1} + (-1)^{k} - 1}{3 \cdot 2^{k+1}} \] Next, re-arrange the numerator: \[ = \frac{2^{(k+1) + 1} + (-1)^{k+1}}{3 \cdot 2^{k+1}} \] The form is now the same as the original given formula for n = k+1. Thus, the formula is true for n = k+1, which completes the inductive step.
04

Conclusion

Since the base case holds and the inductive step has been proven, by mathematical induction, the formula \[ \sum_{j=0}^{n} \left(-\frac{1}{2}\right)^{j} = \frac{2^{n+1} + (-1)^{n}}{3 \cdot 2^{n}} \] is true for all nonnegative integers n.

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Key Concepts

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

Base Case
When using mathematical induction, the first step is always verifying the base case. This establishes that the statement holds true for the smallest possible value of the variable, usually starting at 0 or 1. In our exercise, we start with n=0. We substitute 0 into the given summation formula and both sides must be equal to show that it works.
For the summation part: \( \sum_{j=0}^{0} \left(-\frac{1}{2}\right)^j = \left(-\frac{1}{2}\right)^0 = 1 \)
And the right-hand side of the formula becomes:
\[\frac{2^{0+1}+(-1)^{0}}{3 \cdot 2^{0}} = \frac{2+1}{3 \cdot 1} = \frac{3}{3} = 1 \]
Since both sides are equal, the base case is verified and true. This means our formula holds when n=0, allowing us to proceed to the next step.
Inductive Hypothesis
The second step in mathematical induction involves the inductive hypothesis. Here, we assume that the formula holds true for an arbitrary nonnegative integer \(k\). This assumption is crucial because it forms the foundation for proving the next step.
In our case, we assume that:
\[ \sum_{j=0}^k \left(-\frac{1}{2}\right)^j = \frac{2^{k+1} + (-1)^k}{3 \cdot 2^k} \]
This assumption does not need to be proven; instead, it will be used to prove that the formula is true for \(n = k+1\). By relying on the truth of the statement for \(k\), we aim to show it also holds for the next value, thereby supporting the overall structure of induction.
Summation Formula
The summation formula is the final step where we use the inductive hypothesis to prove that the formula is valid for \(n = k+1\). We start by adding one more term to the left-hand side of the assumption:
\[ \sum_{j=0}^{k+1} \left(-\frac{1}{2}\right)^j = \left(\sum_{j=0}^k \left(-\frac{1}{2}\right)^j \right) + \left(-\frac{1}{2}\right)^{k+1} \]
Next, we replace the summation part with our inductive hypothesis:
\[ \frac{2^{k+1} + (-1)^k}{3 \cdot 2^k} + \left(-\frac{1}{2}\right)^{k+1} \]
By simplifying and finding a common denominator, we get:
\[ \frac{2^{(k+1)+1} + (-1)^{k+1}}{3 \cdot 2^{k+1}} \]
We see that this matches the original formula for \(n = k+1\), completing the inductive proof. Thus, we have shown the formula holds for all nonnegative integers \(n\), thanks to mathematical induction. By carefully combining these steps, induction ensures that the formula starts true and remains true as it progresses through each integer.

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Most popular questions from this chapter

Sometimes we cannot use mathematical induction to prove a result we believe to be true, but we can use mathematical induction to prove a stronger result. Because the inductive hypothesis of the stronger result provides more to work with, this process is called inductive loading. We use inductive loading in Exercise \(74-76\) . Suppose that we want to prove that $$ \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2 n-1}{2 n}<\frac{1}{\sqrt{3 n}} $$ for all positive integers \(n .\) a) Show that if we try to prove this inequality using mathematical induction, the basis step works, but the inductive step fails. b) Show that mathematical induction can be used to prove the stronger inequality $$ \frac{1}{2} \cdot \frac{3}{4} \dots \frac{2 n-1}{2 n}<\frac{1}{\sqrt{3 n+1}} $$ for all integers \(n\) greater than \(1,\) which, together with a verification for the case where \(n=1\) , establishes the weaker inequality we originally tried to prove using mathematical induction.

(Requires calculus) Use mathematical induction to prove that the derivative of \(f(x)=x^{n}\) equals \(n x^{n-1}\) whenever \(n\) is a positive integer. (For the inductive step, use the product rule for derivatives.)

Prove that the program segment $$ \begin{array}{l}{y :=1} \\ {z :=x+y}\end{array} $$ is correct with respect to the initial assertion \(x=0\) and the final assertion \(z=1\)

Prove that in a bit string, the string 01 occurs at most one more time than the string \(10 .\)

Let \(a_{1}, a_{2}, \ldots, a_{n}\) be a list of \(n\) distinct real numbers. How many comparisons are needed to form two sublists from this list, the first containing elements less than \(a_{1}\) and the second containing elements greater than \(a_{1} ?\)
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