Problem 2 - Calculus Single Variable

Question

P In Exercises \(1-10\), write out the sum, and perform the addition. $$ \sum_{j=0}^{6}(2 j-1) $$

Step-by-Step Solution

Answer

The sum is 35.

Step 1: Understand the Exercise

We are given a sum \[ \sum_{j=0}^{6}(2j - 1) \]and we need to write out and calculate it. This represents a series where each term is generated by the expression \(2j - 1\) and \(j\) ranges from 0 to 6.

Step 2: Write Out the Terms of the Sum

Substitute each value of \(j\) from 0 to 6 into the expression \(2j - 1\):- For \(j = 0\): \(2(0) - 1 = -1\)- For \(j = 1\): \(2(1) - 1 = 1\)- For \(j = 2\): \(2(2) - 1 = 3\)- For \(j = 3\): \(2(3) - 1 = 5\)- For \(j = 4\): \(2(4) - 1 = 7\)- For \(j = 5\): \(2(5) - 1 = 9\)- For \(j = 6\): \(2(6) - 1 = 11\)So, the series terms are: \(-1, 1, 3, 5, 7, 9, 11\).

Step 3: Perform the Addition

Add the terms from the series:\[-1 + 1 + 3 + 5 + 7 + 9 + 11\]1. \(-1 + 1 = 0\)2. \(0 + 3 = 3\)3. \(3 + 5 = 8\)4. \(8 + 7 = 15\)5. \(15 + 9 = 24\)6. \(24 + 11 = 35\)Thus, the total sum is 35.

Key Concepts

series

A series is essentially a sum of terms, and it gives us insight into the behavior and properties of numbers and sequences. When you see a summation symbol, it denotes a series where we are adding up individual terms. In mathematical notation, this is often expressed as \( \sum_{j=a}^{b} f(j)\), which means you sum the terms generated by the function \( f(j) \) as \( j \) goes from \( a \) to \( b \).
In our case, the series is given by \( \sum_{j=0}^{6}(2j - 1) \). This tells us that for each integer \( j \) starting from 0 to 6, we plug \( j \) into the expression \( 2j - 1 \), and then add up all those results.

• The terms generated in this example are: -1, 1, 3, 5, 7, 9, 11.
• The sum of these numbers gives us the final result of the series.

Series can be finite or infinite, depending on whether the upper limit \( b \) is a specific number or approaches infinity.

arithmetic progression

An arithmetic progression (or arithmetic sequence) is a sequence of numbers where each term after the first is found by adding a constant difference to the previous term. This constant difference is known as the common difference.
For our series \( -1, 1, 3, 5, 7, 9, 11 \), you can easily notice that each term after the initial \( -1 \) is obtained by adding 2 to the previous term.

• The common difference here is 2, calculated by subtracting consecutive terms, such as \( 1 - (-1) = 2 \).
• An arithmetic progression is a simple way to form a series of numbers that increase or decrease steadily.

If you are given the first term \( a \) and the common difference \( d \), you can find any term in an arithmetic sequence using the formula:
\[ a_n = a + (n-1)d \]
Here, \( a \) is \(-1\), \( d \) is \(2\), and \( n \) denotes the term's position in the sequence.

addition

Addition is one of the basic arithmetic operations and is essential in working with series and sequences. It involves calculating the total or sum by combining two or more numbers. In the step-by-step solution of the exercise provided, we are tasked with adding together the terms of the series:
\[-1 + 1 + 3 + 5 + 7 + 9 + 11 \]
This is done by sequentially adding the numbers in the order they appear:

• First, add \(-1 + 1\), which equals \(0\).
• Next, take the resulting \(0\) and add \(3\) to get \(3\).
• Continue this process, each time adding the next number in the series.

Addition allows us to find the cumulative value of the series step by step. It requires focusing and carefully carrying out each operation, as missing a step can lead to incorrect results. By understanding each individual addition, you gain solid insight into how summation works intuitively.