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a) If \(a_{0}, a_{1}, a_{2}, a_{3}\) is a list of four real numbers, what is \(\sum_{i=1}^{3}\left(a_{t}-a_{i-1}\right) ?\) b) Given a list- \(a_{0}, a_{1}, a_{2}, \ldots, a_{n}-\) of \(n+1\) real numbers, where \(n\) is a positive integer, determine \(\sum_{i=1}^{n}\left(a_{i}-a_{i-1}\right)\). c) Determine the value of \(\sum_{i=1}^{100}\left(\frac{1}{1+2}-\frac{1}{i+1}\right)\).

Short Answer

Expert verified
a) The sum simplifies to \(a_3 - a_0\). b) The sum simplifies to \(a_n - a_0\). c) The sum simplifies to \(\frac{100}{3} - (\frac{1}{2} + \frac{1}{3} +...+ \frac{1}{101})\).

Step by step solution

01

Solve for Part (a)

To solve, we simply substitute into our formula \(\sum_{i=1}^{3}\left(a_{i}-a_{i-1}\right)\). The calculated expression becomes: \((a_1-a_0) + (a_2-a_1) +(a_3-a_2)\). Now, some of the terms cancel out and the expression simplifies to \(a_3-a_0\).
02

Solve for Part (b)

By using the same approach as in part (a), we substitute into our equation \(\sum_{i=1}^{n}\left(a_{i}-a_{i-1}\right)\). The expression becomes: \((a_1-a_0) + (a_2-a_1) + . . . +(a_n-a_{n-1})\). Again, terms cancel out and the expression simplifies to \(a_n - a_0\).
03

Solve for Part (c)

In \(\sum_{i=1}^{100}\left(\frac{1}{1+2}-\frac{1}{i+1}\right)\), the term \(\frac{1}{1+2}\) is a constant and it doesn’t affect the pattern of the sum: \(\sum_{i=1}^{100}\frac{1}{1+2}\) simplifies to \(\frac{100}{3}\). For the second term \(\sum_{i=1}^{100}\left(-\frac{1}{i+1}\right)\), it simplifies to \(-\frac{1}{2} - \frac{1}{3} - \ldots - \frac{1}{101}\). The sum of both expressions gives the final answer.
04

Simplify Part (c)

On simplifying further, the final expression becomes \(\frac{100}{3} - (\frac{1}{2} + \frac{1}{3} +...+ \frac{1}{101})\). Note that this results in a series sum of fractions, so the exact result might be cumbersome to calculate by hand, but it's possible with calculators or software.

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Key Concepts

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

Series and Sequence
When we talk about series and sequences in mathematics, we refer to two distinct yet closely related concepts. A sequence is simply an ordered list of numbers. Each number in this list is called a term. For example, the sequence 2, 4, 6, 8 is an ordered list of even numbers. An important point to note is that sequences can be finite or infinite. - A finite sequence has a limited number of terms. - An infinite sequence goes on indefinitely.
On the other hand, a series is what you get when you add up the terms of a sequence. Using the previous example, the sum 2 + 4 + 6 + 8 is a series. In mathematics, we often use a capital sigma symbol, \( \Sigma \), to represent such sums. What's interesting about sequences and series is how they form the foundation for many mathematical concepts, including calculus and algebra.
Whether you're learning arithmetic series, geometric, or other types, mastering sequences and series gives you a powerful tool for analyzing mathematical patterns!
Arithmetic Progression
Arithmetic Progression, often abbreviated as AP, is a special type of sequence. In an AP, each term after the first is obtained by adding a constant value to the previous term. This constant is known as the 'common difference.' - For example, consider the sequence 5, 7, 9, 11, which has a common difference of 2.
The formula for finding the \( n \)-th term of an arithmetic progression is: \[ a_n = a_1 + (n-1) imes d \] where \( a_1 \) is the first term, \( n \) is the term number, and \( d \) is the common difference.
In the exercise, the calculation of sums involving terms like \( a_i - a_{i-1} \) follows the idea of eliminating middle terms. This is akin to a telescoping sum, which is simplified through cancellation, eventually leading to an arithmetic pattern that helps in solving such problems. Understanding arithmetic progressions in such contexts can simplify many summation problems!
Telescoping Series
A telescoping series is a sequence where many of the terms cancel each other out when the terms are summed. This concept takes its name from the collapsing feature of a telescope, where parts slide into each other so that only the endpoints remain.
Consider the sum from the exercise: \( \sum_{i=1}^{n}(a_i-a_{i-1}) \). If you expand the terms, you see a pattern: - \( (a_1-a_0) + (a_2-a_1) + (a_3-a_2) + ... + (a_n-a_{n-1}) \)
Notice that most terms cancel each other out, leaving only \( a_n - a_0 \). This type of cancellation makes evaluating telescoping series often simpler than they appear.
Telescoping series are not only useful in arithmetic progressions but appear across different areas of mathematics, providing elegant solutions to otherwise complex problems.
Real Numbers
Real numbers make up one of the most critical sets in mathematics. They include all the numbers that can be found on the number line. This means they encompass: - Whole numbers (like 1, 2, 3), - Fractions and decimals (like \( \frac{1}{2} \) and 0.5), - Irrational numbers (like \( \sqrt{2} \) and \( \pi \)) as well.
Real numbers provide a way to represent continuous quantities and are vital in calculus, algebra, measurement, and much more. Mathematical operations such as adding, subtracting, multiplying, or dividing real numbers follow familiar rules that you learn early on.
In the given exercise, terms like \( a_0, a_1, a_2 \) are real numbers. When handling sums or sequences, understanding the properties of real numbers ensures precise calculations and the transition between different mathematical domains. Essentially, real numbers form the backbone of most mathematical computations you'll encounter from basic algebra to advanced calculus.

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Most popular questions from this chapter

For any given set in a tennis tournament, opponent A can beat opponent B in seven different ways. (At 6-6 they play a tie breaker.) The first opponent to win three sets wins the tournament. (a) In how many ways can scores be recorded with \(A\) winning in five sets? (b) In how many ways can scores be recorded with the tournament requiring at least four sets?

In ordering the daily special at a diner, a customer has a choice of three entrees and may select any two of six available vegetables. a) How many different dinners can she select if (i) she must select two different vegetables? (ii) she is permitted to have two helpings of the same vegetable? b) Answer parts (i) and (ii) of part (a) if she also has a choice of tomato juice, orange juice, or bean soup as an appetizer.

Consider the following Pascal program segment, where \(i, j, k, n\), and counter are integer variables. Earlier in the program, the user is prompted for a positive integer establishing the value of \(n\) for that particular run of the program. counter : \(=0 ;\) For \(i:=1\) to \(\mathrm{n}\) do For \(j:=1\) to 1 do For \(k:=1\) to j do counter : \(=\) counter \(+1 ;\) We shall determine, in two different ways, the number of times the statement $$ \text { counter : = counter }+1 $$ is executed. (This is also the value of counter after execution of the program segment.) From the result in Example 1.37, we know that the statement is executed \(\left({ }^{n+3}{ }^{-1}\right)=\left({ }^{n+2}{ }^{2}\right)\) times. For a fixed value of \(i\), the For loops involving \(j\) and \(k\) result in \(\left(\begin{array}{c}l+1 \\ 2\end{array}\right)\) executions of the counter increment statement. Consequently, \(\left(\begin{array}{c}m+2 \\\ 3\end{array}\right)=\sum_{i=1}^{n}\left(\begin{array}{c}i+1 \\\ 2\end{array}\right)\). Use this result to obtain a summation formula for \(1^{2}+2^{2}+3^{2}+\cdots+n^{2}=\sum_{i=1}^{n} i^{2} .\)

a) In how many ways can we select five coins from a collection of 10 consisting of one penny, one nickel, one dime, one quarter, one half-dollar, and five (identical) Susan B. Anthony dollars? b) In how many ways can we select \(n\) objects from a collection of size \(2 n\) that consists of \(n\) distinct and \(n\) identical objects?

Write a computer program (or develop an algorithm) to compute the integer solutions for $$ x_{1}+x_{2}+x_{3}=10, \quad 0 \leq x_{i}, \quad 1 \leq i \leq 3 $$

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