Problem 14 - Algebra 2 and Trigonometry

Question

In \(3-14 :\) a. Write each arithmetic series as the sum of terms. b. Find each sum. $$ \sum_{n=0}^{5}(-2 n)^{n+1} $$

Step-by-Step Solution

Answer

The sum of the series is 968.

Step 1: Understand the Series Notation

The given expression is \( \sum_{n=0}^{5}(-2n)^{n+1} \). This means we need to calculate each term from \( n = 0 \) to \( n = 5 \) using the expression \((-2n)^{n+1}\).

Step 2: Calculate Each Term in the Series

For \( n = 0 \), the term is \((-2 \times 0)^{0+1} = 0^1 = 0\).For \( n = 1 \), the term is \((-2 \times 1)^{1+1} = (-2)^2 = 4\).For \( n = 2 \), the term is \((-2 \times 2)^{2+1} = (-4)^3 = -64\).For \( n = 3 \), the term is \((-2 \times 3)^{3+1} = (-6)^4 = 1296\).For \( n = 4 \), the term is \((-2 \times 4)^{4+1} = (-8)^5 = -32768\).For \( n = 5 \), the term is \((-2 \times 5)^{5+1} = (-10)^6 = 1000000\).

Step 3: Write the Series as a Sum

Using the calculated terms, the series can be written as:\( 0 + 4 - 64 + 1296 - 32768 + 1000000 \).

Step 4: Find the Sum of the Series

Sum the terms from Step 3:\( 0 + 4 - 64 + 1296 - 32768 + 1000000 = 968\).

Key Concepts

Understanding Sigma Notation

Sigma notation, represented by the symbol \( \Sigma \), allows us to write long sums in a compact form. This notation elegantly expresses the idea of adding a sequence of terms. The notation \( \sum_{n=a}^{b}f(n) \) means that we need to compute the sum of the function \( f(n) \) as \( n \) takes on values from \( a \) to \( b \). Each term is calculated by substituting \( n \) into the function \( f(n) \).
In our exercise, \( \sum_{n=0}^{5}(-2n)^{n+1} \), we are tasked with finding the sum by evaluating \((-2n)^{n+1}\) for each integer \( n \) from 0 to 5. This formula within the sigma notation indicates how each term in the series is formed.

The purpose of sigma notation is not just brevity, but clarity. It helps understand the pattern or rule governing the series at a glance. This consolidated view is especially helpful for more complex calculations, where manually writing out each step would be cumbersome.

• Remember, the index \( n \) indicates the position in the sequence, and by substituting the index into the expression, you discover each specific term in the series.

Calculating the Sum of a Series

The sum of a series involves finding the total that results from adding all the individual terms of a sequence. Specifically, in an arithmetic series, each term is derived following a particular rule or pattern and then summed up. To calculate the sum as seen in our example, we break down each term, evaluate it, and then add them.
The process for finding the sum of the series \( \sum_{n=0}^{5}(-2n)^{n+1} \) involves calculating terms for each \( n \) value, individually substituting \( n\):

• For \( n = 0 \), the term is \( 0 \).
• For \( n = 1 \), the term is \( 4 \).
• For \( n = 2 \), the term is \( -64 \).
• For \( n = 3 \), the term is \( 1296 \).
• For \( n = 4 \), the term is \( -32768 \).
• For \( n = 5 \), the term is \( 1000000 \).

The sum of these terms is a straightforward addition: \( 0 + 4 - 64 + 1296 - 32768 + 1000000 = 968 \).
This approach, though seemingly tedious, ensures accuracy by breaking the task into manageable steps.

Evaluating Mathematical Expressions

Evaluating mathematical expressions in a series can be the trickiest part, as it requires handling operations such as exponentiation and multiplication. In our example, expression evaluation requires careful substitution and computation.
For each term derived from the expression \((-2n)^{n+1}\), follow these steps:

• Multiply \( n \) by \(-2\) as the first operation.
• Raise the result to the power \( n+1 \), which depends on the current value of \( n \).

For example, substituting \( n=3 \) into \((-2n)^{n+1}\), you multiply: \(-2 \times 3 = -6\), and then raise this to the fourth power: \((-6)^4 = 1296\).
Each operation must be executed in sequence, respecting the order of operations (parentheses and exponents first, followed by multiplication).
It's helpful to write down intermediate steps, especially when dealing with large numbers or more complex expressions. This practice prevents errors and clarifies each step's role in forming the final series sum.