Ch. 8 - Sequences, Induction, and Probability

Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook

Blitzer 8th Edition

Ch. 8 - Sequences, Induction, and Probability

Problem 23

Chapter 9, Problem 23

Use mathematical induction to prove that each statement is true for every positive integer n. 1/(1 · 2) + 1/(2 · 3) + 1/(3 · 4) + ... + 1/(n(n+1)) = n/(n + 1)

Verified step by step guidance

Start by defining the statement to prove using mathematical induction. Let $ P(n) $ be the statement: \[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \cdots + \frac{1}{n(n+1)} = \frac{n}{n+1} \].

Check the base case $ n=1 $. Substitute $ n=1 $ into the left-hand side and right-hand side of the equation to verify that both sides are equal.

Assume the statement $ P(k) $ is true for some positive integer $ k $, that is, assume \[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} = \frac{k}{k+1} \]. This is the induction hypothesis.

Use the induction hypothesis to prove $ P(k+1) $ is true. Add the next term $ \frac{1}{(k+1)(k+2)} $ to both sides of the assumed equation and simplify the right-hand side to show it equals $ \frac{k+1}{k+2} $.

Conclude that since the base case is true and the induction step holds, by mathematical induction, the statement is true for every positive integer $ n $.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Mathematical Induction

Mathematical induction is a proof technique used to establish that a statement holds for all positive integers. It involves two steps: proving the base case (usually n=1) is true, and then proving that if the statement holds for an arbitrary integer k, it also holds for k+1. This creates a chain of truth for all n.

Telescoping Series

A telescoping series is a sum where many terms cancel out when expanded, simplifying the expression significantly. In this problem, the sum of fractions 1/(k(k+1)) can be decomposed into partial fractions, allowing terms to cancel and making it easier to find a closed form for the sum.

Partial Fraction Decomposition

Partial fraction decomposition breaks a complex rational expression into simpler fractions that are easier to sum or integrate. For example, 1/(k(k+1)) can be written as 1/k - 1/(k+1), which helps in simplifying the sum and proving the formula using induction.

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