Ch. 5 - Systems of Equations and Inequalities

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Blitzer 8th Edition

Ch. 5 - Systems of Equations and Inequalities

Problema 51

Capitolo 6, Problema 51

Find the partial fraction decomposition for 1/x(x+1) and use the result to find the following sum:
$Mathematical series showing partial fractions: 1/1·2 + 1/2·3 + ... + 1/99·100.$

Guida verificata passo dopo passo

Start by expressing the general term of the series as a partial fraction. The general term is $ \frac{1}{n(n+1)} $. Set up the partial fraction decomposition as $ \frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1} $.

Multiply both sides of the equation by $ n(n+1) $ to clear the denominators: $ 1 = A(n+1) + Bn $.

Expand the right side: $ 1 = An + A + Bn = (A + B)n + A $.

Equate the coefficients of like terms on both sides. Since the left side is a constant 1, the coefficient of $ n $ on the right must be 0, and the constant term must be 1. This gives the system: $ A + B = 0 $ and $ A = 1 $.

Solve the system to find $ A = 1 $ and $ B = -1 $. Therefore, the partial fraction decomposition is $ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} $. Use this to rewrite each term in the sum and observe the telescoping pattern to simplify the sum.

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Partial Fraction Decomposition

Partial fraction decomposition is a method used to express a rational function as a sum of simpler fractions. For example, the fraction 1/[x(x+1)] can be decomposed into A/x + B/(x+1), where A and B are constants found by solving equations. This technique simplifies complex fractions and is useful for summation and integration.

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Decomposition of Functions

Telescoping Series

A telescoping series is a series whose partial sums simplify by canceling intermediate terms, leaving only the first and last terms. When partial fractions are applied to terms like 1/(n(n+1)), the series often telescopes, making it easier to find the sum of many terms without adding each individually.

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Summation of Series

Summation of series involves finding the total sum of a sequence of terms. Using partial fraction decomposition and recognizing telescoping behavior allows for efficient calculation of sums like 1/(1·2) + 1/(2·3) + ... + 1/(99·100), by reducing the problem to evaluating a few terms rather than all 99.

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