Problem 8 - Precalculus: Mathematics for Calculus

Question

Find the first four terms and the 100 th term of the sequence whose \(n\)th term is given. \(a_{n}=\left(\frac{-1}{3}\right)^{n}\)

Step-by-Step Solution

Answer

First four terms: \(\frac{-1}{3}, \frac{1}{9}, \frac{-1}{27}, \frac{1}{81}\); 100th term: \(\frac{1}{3^{100}}\).

Step 1: Understand the Sequence Formula

The sequence is given by the formula \(a_n = \left( \frac{-1}{3} \right)^n\). This means each term is obtained by raising \( \frac{-1}{3} \) to the power of \(n\). Here, \(n\) is the position of the term in the sequence.

Step 2: Calculate the First Term (n=1)

To find the first term, substitute \(n = 1\) into the formula: \[a_1 = \left( \frac{-1}{3} \right)^1 = \frac{-1}{3}.\] Hence, the first term is \(\frac{-1}{3}\).

Step 3: Calculate the Second Term (n=2)

For the second term, substitute \(n = 2\) into the formula: \[a_2 = \left( \frac{-1}{3} \right)^2 = \left( \frac{1}{9} \right).\] Therefore, the second term is \(\frac{1}{9}\).

Step 4: Calculate the Third Term (n=3)

Substitute \(n = 3\) into the formula to find the third term: \[a_3 = \left( \frac{-1}{3} \right)^3 = \left( -\frac{1}{27} \right).\] Thus, the third term is \(-\frac{1}{27}\).

Step 5: Calculate the Fourth Term (n=4)

To find the fourth term, substitute \(n = 4\) into the formula: \[a_4 = \left( \frac{-1}{3} \right)^4 = \left( \frac{1}{81} \right).\] Hence, the fourth term is \(\frac{1}{81}\).

Step 6: Calculate the 100th Term (n=100)

Finally, substitute \(n = 100\) into the formula to find the 100th term: \[a_{100} = \left( \frac{-1}{3} \right)^{100} = \left( \frac{1}{3^{100}} \right).\] This results in a positive very small fraction \(\frac{1}{3^{100}}\).

Key Concepts

Sequence Formula

The sequence formula is a mathematical expression that defines the relationship between the position of a term in a sequence and its value. In the original exercise, we are given the formula \(a_n = \left( \frac{-1}{3} \right)^n\). This equation tells us that each term \(a_n\) is formed by raising \(\frac{-1}{3}\) to the power of \(n\), where \(n\) is the term's position in the sequence.

• It is crucial to understand that this formula governs the entire sequence.
• Each term is related to its position through the operation of exponentiation.

Understanding this formula allows you to calculate any term in the sequence by substituting the corresponding value of \(n\) into the formula. Keep in mind that the sign and magnitude of the fraction change with the power of \(n\). This is an essential concept to grasp when working with sequences of this type.

Exponentiation

Exponentiation is a key mathematical operation in this problem where a number is multiplied by itself a given number of times. When you see a power in an arithmetic sequence like \(\left( \frac{-1}{3} \right)^n\), exponentiation determines the value of each term in the series.

• For \(n = 1\), the term is \(\left( \frac{-1}{3} \right)^1 = \frac{-1}{3}\).
• For \(n = 2\), you perform \(\left( \frac{-1}{3} \right) \times \left( \frac{-1}{3} \right) = \frac{1}{9}\), as the negative sign squared becomes positive.
• For \(n = 3\), \(\left( \frac{-1}{3} \right)^3 = \frac{-1}{3} \times \frac{1}{9} = -\frac{1}{27}\), switches the sign back to negative.

The alternating sign feature of this sequence highlights an important concept of exponentiation: odd powers of negative numbers result in negative values, whereas even powers result in positive values. This alternation significantly impacts the nature of the sequence and is integral to understanding how the sequence evolves.

Term Calculation

Calculating terms in an arithmetic sequence involves substituting the desired term's position into the sequence formula. Let's see how each term is computed step-by-step from the original exercise:

• **First Term**: At \(n=1\), substitute to get \(a_1 = \left( \frac{-1}{3} \right)^1 = \frac{-1}{3}\).
• **Second Term**: At \(n=2\), substituting gives \(a_2 = \left( \frac{-1}{3} \right)^2 = \frac{1}{9}\).
• **Third Term**: At \(n=3\), calculation results in \(a_3 = \left( \frac{-1}{3} \right)^3 = -\frac{1}{27}\).
• **Fourth Term**: At \(n=4\), find that \(a_4 = \left( \frac{-1}{3} \right)^4 = \frac{1}{81}\).
• **100th Term**: A large power such as \(n=100\) results in a very small positive fraction, \(a_{100} = \frac{1}{3^{100}}\).

This methodical substitution process is straightforward and can be applied to determine any specific term desired. Being systematic in term calculations ensures accuracy, especially when working with high exponent values, which simplify the negative power alternation by producing positive results.