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Chapter 13: Problem 12

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ \frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\cdots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1} $$

Short Answer

Expert verified
The given statement is true for all natural numbers \(n\) by mathematical induction.

Step by step solution

01

- Basis Step

Verify the statement for the initial value of the natural number, usually for \(n = 1\). In this case, the left-hand side (LHS) for \(n = 1\) is: \[ \frac{1}{1 \times 3} = \frac{1}{3} \] The right-hand side (RHS) for \(n = 1\) is: \[ \frac{1}{2 \times 1 + 1} = \frac{1}{3} \] Since LHS = RHS, the statement holds for \(n = 1\).
02

- Inductive Step

Assume the statement is true for some natural number \(k\), that is, assume: \[ \frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \text{...} + \frac{1}{(2k - 1)(2k + 1)} = \frac{k}{2k + 1} \] This is called the induction hypothesis.
03

- Inductive Step (Prove for \(k+1\))

Prove the statement for \(k + 1\) based on the induction hypothesis. This means we need to show: \[ \frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \text{...} + \frac{1}{(2k - 1)(2k + 1)} + \frac{1}{(2(k + 1) - 1)(2(k + 1) + 1)} = \frac{k + 1}{2(k + 1) + 1} \] Using the induction hypothesis, the LHS becomes: \[ \frac{k}{2k + 1} + \frac{1}{(2k + 1)(2k + 3)} \]
04

- Simplify the Expression

Combine the fractions on the LHS as follows: \[ \frac{k}{2k + 1} + \frac{1}{(2k + 1)(2k + 3)} = \frac{k(2k + 3) + 1}{(2k + 1)(2k + 3)} \] Simplify the numerator: \[ \frac{2k^2 + 3k + 1}{(2k + 1)(2k + 3)} \] Observe that the numerator can be factored as: \[ 2k^2 + 3k + 1 = (k + 1)(2k + 1) \] Therefore, the LHS simplifies to: \[ \frac{(k + 1)(2k + 1)}{(2k + 1)(2k + 3)} = \frac{k + 1}{2k + 3} \] This matches the RHS for \(n = k + 1\).
05

- Conclusion

By mathematical induction, since the basis step is true and the inductive step holds, the statement: \[ \frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \text{...} + \frac{1}{(2n - 1)(2n + 1)} = \frac{n}{2n + 1} \] is true for all natural numbers \(n\).

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Key Concepts

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

Mathematical Induction
Mathematical induction is a technique used to prove that a statement is true for all natural numbers. It consists of two main steps: the basis step and the inductive step. In the basis step, we verify the statement for the smallest natural number, usually 1. Once this is established, the inductive step assumes the statement is true for some random natural number and then proves it for the next number. This method is powerful and widely used in proving formulas and equations concerning natural numbers.
Natural Numbers
Natural numbers are the set of positive integers starting from 1, 2, 3, and so on. In mathematical induction, these numbers form the basis for proving that a statement holds for all elements within this set. They are crucial because the process of induction relies on establishing truth sequentially through these numbers. Natural numbers are used because they have a clear starting point (1) and progress in a way that allows for step-by-step verification.
Inductive Hypothesis
The inductive hypothesis is an assumption made during the inductive step of mathematical induction. We assume that the statement is true for a particular natural number, let鈥檚 say k. This hypothesis is then used to prove that the statement holds for the next number, k+1. It鈥檚 a pivotal part of the proof because it bridges the gap between the known truth (basis step) and the desired proof for all natural numbers. By demonstrating that if it's true for k, it's true for k+1, the induction hypothesis validates the statement for an infinite set of natural numbers.
Algebraic Proof
Algebraic proof involves manipulating algebraic expressions to show that both sides of an equation are identical. In the context of mathematical induction, this often means starting with the induction hypothesis and adding the next term in the sequence. By simplifying and combining terms, we show that the extended equation holds true, which completes the inductive step. For example, in the given exercise, combining fractions and simplifying them algebraically helps to demonstrate that the statement holds for k+1, thus cementing the proof for all natural numbers.

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Most popular questions from this chapter

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Find a rectangular equation of the plane curve with parametric equations \(x(t)=t+5\) and \(y(t)=\sqrt{t}\) for \(t \geq 0\).

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Challenge Problem Paper Creases If a sheet of paper is folded in half by folding the top edge down to the bottom edge, one crease will result. If the folded paper is folded in the same manner, the result is three creases. With each fold, the number of creases can be defined recursively by \(c_{1}=1, c_{n+1}=2 c_{n}+1\) (a) Find the number of creases for \(n=3\) and \(n=4\) folds. (b) Use the given information and your results from part (a) to find a formula for the number of creases after \(n\) folds, \(c_{n}\), in terms of the number of folds alone. (c) Use the Principle of Mathematical Induction to prove that the formula found in part (b) is correct for all natural numbers. (d) Tosa Tengujo is reportedly the world's thinnest paper with a thickness of \(0.02 \mathrm{~mm}\). If a piece of this paper could be folded 25 times, how tall would the stack be?

If \(f(x)=x^{2}-6\) and \(g(x)=\sqrt{x+2},\) find \(g(f(x))\) and state its domain.

\(\sqrt{89}\)
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