/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 74 Use the summation properties and... [FREE SOLUTION] | 91Ó°ÊÓ

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Use the summation properties and rules to evaluate each series. $$\sum_{i=1}^{6}\left(2+i-i^{2}\right)$$

Short Answer

Expert verified
-58

Step by step solution

01

- Expand and Separate the Summation

Use the linearity property of summations to separate the terms inside the summation oindent\[\sum_{i=1}^{6} \left( 2 + i - i^2 \right) = \sum_{i=1}^{6} 2 + \sum_{i=1}^{6} i - \sum_{i=1}^{6} i^2\]
02

- Evaluate the Summation for 2

The summation of 2 from 1 to 6 is oindent\[\sum_{i=1}^{6} 2 = 2 \times 6 = 12\]
03

- Evaluate the Summation for i

The sum of the first n natural numbers is given by the formula: \(\sum_{i=1}^{n} i = \frac{n(n+1)}{2}\). So for this problem:oindent\[\sum_{i=1}^{6} i = \frac{6 \times (6 + 1)}{2} = \frac{6 \times 7}{2} = 21\]
04

- Evaluate the Summation for i^2

The sum of the squares of the first n natural numbers is given by the formula: \(\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\). So for this problem:oindent\[\sum_{i=1}^{6} i^2 = \frac{6 \times 7 \times 13}{6} = 91\]
05

- Combine the Results

Combine the results from the previous steps:oindent\[\sum_{i=1}^{6} \left( 2 + i - i^2 \right) = 12 + 21 - 91 = -58\]

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Key Concepts

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

series evaluation
Series evaluation is the process of calculating the sum of a sequence of numbers, often represented by a summation. In this exercise, we evaluate the series \(\sum_{i=1}^{6} (2 + i - i^2)\) using summation properties. The linearity of summation allows us to separate the series into simpler parts: \[\sum_{i=1}^{6} (2 + i - i^2) = \sum_{i=1}^{6} 2 + \sum_{i=1}^{6} i - \sum_{i=1}^{6} i^2 \]. Breaking down the series:
  • The first part is the summation of a constant, 2.
  • The second part is the summation of natural numbers, represented by i.
  • The third part involves the summation of squared terms, i².
These can be simplified and evaluated individually, then combined to get the final result. Each step relies on known summation formulas to ensure accurate and efficient calculation.
natural numbers
Natural numbers are positive integers starting from 1 and moving upwards (1, 2, 3,...). They form the basis of many mathematical operations and are crucial in evaluating series. In this problem, the summation of natural numbers \(\sum_{i=1}^{n} i\) is calculated using the formula \[\sum_{i=1}^{n} i = \frac{n(n+1)}{2}\]. Applying this to our series where n = 6: \[\sum_{i=1}^{6} i = \frac{6 \times (6 + 1)}{2} = 21\]. Understanding this formula is essential because it simplifies the process of adding up natural numbers within any range, ensuring you can quickly find the sum without manual counting.
sum of squares
The sum of squares concept deals with summing the squares of natural numbers up to a specific limit. The summation of squares formula is given as \[\sum_{i=1}^{n} i^{2} = \frac{n(n+1)(2n+1)}{6}\]. This is handy when evaluating series with squared terms. For our problem: \[\sum_{i=1}^{6} i^2 = \frac{6 \times 7 \times 13}{6} = 91\]. By using this formula, you can efficiently find the sum of squares for any number range, saving time and reducing errors. Summing squared terms is common in various mathematical contexts, making this a valuable tool in your problem-solving toolkit.

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Most popular questions from this chapter

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