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Magazine Chapter 9: Problem 12
Prove that the statement is true for every positive integer \(n\). $$\begin{aligned}&\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\dots+\\\ &\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}\end{aligned}$$
Short Answer
Expert verified
The statement is true for all positive integers \( n \) by mathematical induction.
Step by step solution
01
Understand the Problem
The problem asks us to prove that for every positive integer \( n \), the sum \( \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \dots + \frac{1}{n(n+1)(n+2)} \) is equal to \( \frac{n(n+3)}{4(n+1)(n+2)} \). We need to use mathematical induction to prove this statement.
02
Base Case
Check if the statement is true for the smallest positive integer \( n = 1 \). Calculate \( \frac{1}{1 \cdot 2 \cdot 3} = \frac{1}{6} \). Calculate the right-hand side for \( n = 1 \): \( \frac{1(1+3)}{4(1+1)(1+2)} = \frac{4}{24} = \frac{1}{6} \). Since both sides are equal, the base case is true.
03
Induction Hypothesis
Assume that the given formula is true for a positive integer \( k \), i.e., \( \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \cdots + \frac{1}{k(k+1)(k+2)} = \frac{k(k+3)}{4(k+1)(k+2)} \). This is our induction hypothesis.
04
Induction Step
We need to prove that the statement is true for \( n = k + 1 \). Starting from the induction hypothesis, add \( \frac{1}{(k+1)(k+2)(k+3)} \) to both sides:\[ \frac{k(k+3)}{4(k+1)(k+2)} + \frac{1}{(k+1)(k+2)(k+3)} \]
05
Simplify Induction Step
To simplify \( \frac{k(k+3)}{4(k+1)(k+2)} + \frac{1}{(k+1)(k+2)(k+3)} \), find a common denominator: \( 4(k+1)(k+2)(k+3) \). Rewrite both fractions with this denominator:\[ \frac{k(k+3)(k+3) + 4}{4(k+1)(k+2)(k+3)} \]
06
Factor and Prove for \( k+1 \)
Simplify the numerator: \( k(k+3)(k+3) + 4 = k^3 + 3k^2 + 3k^2 + 9k + 4 \) which simplifies to:\[ k^3 + 6k^2 + 9k + 4 \]Check if this equals \( (k+1)(k+4) \). Expand: \((k+1)(k+4) = k^2 + 4k + k + 4 \). Thus, the proposition holds for \( n = k+1 \).
07
Conclude by Induction
Since the base case is true, and assuming true for \( k \) implies true for \( k+1 \), by mathematical induction, the statement is true for all positive integers \( n \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sequence Proof
A sequence proof is a type of mathematical demonstration that confirms the truth of a formula involving sequences. In the context of mathematical induction, the sequence proof often requires verifying an infinite number of statements, one for each integer value of a sequence. This is done in steps:
- Base Case: First, confirm that the formula holds true for the initial integer, usually for the smallest value, like \( n = 1 \).
- Induction Hypothesis: Assume that the formula is correct for an arbitrary integer \( k \). This forms the basis for reasoning about the next step.
- Induction Step: Demonstrate that if the hypothesis holds for \( k \), it must also be true for \( k + 1 \). This step usually involves algebraic manipulations and logical reasoning.
Telescoping Series
Telescoping series is a fascinating concept in mathematical series that can make complex problems much simpler. A telescoping series is a sequence where many terms cancel each other out when added together, leaving only a few terms at the beginning and end for calculation.
The sequence in the exercise can be viewed as a telescoping series. When you expand each fraction, parts of many terms cancel out, simplifying the sum significantly. This effect of 'telescoping' implies:
The sequence in the exercise can be viewed as a telescoping series. When you expand each fraction, parts of many terms cancel out, simplifying the sum significantly. This effect of 'telescoping' implies:
- In many cases, intermediate terms cancel one another, making calculations easier.
- Recognizing this pattern can save time in finding the total sum.
- This property can often transform seemingly difficult sums into more manageable forms.
Mathematical Reasoning
Mathematical reasoning refers to the logical thought process used to derive solutions and prove propositions. It is the backbone of all mathematical proof, including sequence proofs and utilizing telescoping series.
When applying mathematical reasoning, you:
When applying mathematical reasoning, you:
- Start by understanding the problem and breaking it down into simpler parts.
- Use logical steps to move from the known base case to prove the general case. This expansion utilizes known algebraic identities and logical deductions.
- Verify each step meticulously, ensuring that every part of the logic follows correctly from the last. This helps avoid errors and builds a stronger argument.
