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Chapter 11: Problem 65

Write out and evaluate each sum. $$ \sum_{k=3}^{5} \frac{(-1)^{k}}{k(k+1)} $$

Short Answer

Expert verified
\( \frac{-1}{15} \)

Step by step solution

01

Write the General Term

Identify the general term for the given summation: \ \( \frac{(-1)^k}{k(k+1)} \).
02

Determine the Range of Summation

The summation runs from \( k = 3 \) to \( k = 5 \).
03

Substitute Values of k and Compute Each Term

Calculate each term by substituting values of \( k \): \ When \( k = 3 \): \( \frac{(-1)^3}{3(3+1)} = \frac{-1}{12} \) \ When \( k = 4 \): \( \frac{(-1)^4}{4(4+1)} = \frac{1}{20} \) \ When \( k = 5 \): \( \frac{(-1)^5}{5(5+1)} = \frac{-1}{30} \)
04

Add the Results

Add the computed terms to find the total sum: \ \( \frac{-1}{12} + \frac{1}{20} + \frac{-1}{30} \)
05

Find a Common Denominator

Convert the fractions to have a common denominator: \ \( \frac{-5}{60} + \frac{3}{60} + \frac{-2}{60} = \frac{-4}{60} = \frac{-1}{15} \)

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Key Concepts

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

General Term
In sigma notation, the general term represents the expression that changes based on the index of summation, usually denoted as 'k'. For the given problem, the general term is identified as \( \frac{(-1)^k}{k(k+1)} \). This term tells us how each part of the summation sequence is structured. It incorporates both descriptive elements (e.g., (-1)^k, alternating signs) and variables that vary with 'k', making it pivotal to understand this before diving into the calculations.
Range of Summation
The range of summation lays out the limits within which the index (k) moves. For the above problem, the range is from \( k = 3 \) to \( k = 5 \). This means we will substitute each integer value of k within this range into our general term to compute different terms in the summation.
Substitution
Substitution is the act of replacing the variable 'k' with specific values from the defined range to compute each term in the sequence. Let's break down the substitutions for k = 3, 4, and 5:
  • For \( k = 3 \) : \( \frac{(-1)^3}{3(3+1)} = \frac{-1}{12} \)
  • For \( k = 4 \) : \( \frac{(-1)^4}{4(4+1)} = \frac{1}{20} \)
  • For \( k = 5 \) : \( \frac{(-1)^5}{5(5+1)} = \frac{-1}{30} \)
Common Denominator
Finding a common denominator is essential when adding or subtracting fractions. It allows us to combine the fractions into a single term. For the fractions \( \frac{-1}{12}, \frac{1}{20}, \frac{-1}{30} \), the least common denominator is 60. Converting each fraction, we get:
  • \( \frac{-1}{12} \to \frac{-5}{60} \)
  • \( \frac{1}{20} \to \frac{3}{60} \)
  • \( \frac{-1}{30} \to \frac{-2}{60} \)
Adding these fractions: \( \frac{-5}{60} + \frac{3}{60} + \frac{-2}{60} = \frac{-4}{60} = \frac{-1}{15} \). Now we have the final evaluated sum.

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Most popular questions from this chapter

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