Problem 43 - Precalculus: Mathematics for Calculus

Question

Find the first four partial sums and the \(n\) th partial sum of the sequence \(a_{n}.\) \(a_{n}=\frac{2}{3^{n}}\)

Step-by-Step Solution

Answer

The first four partial sums are \(\frac{2}{3}\), \(\frac{8}{9}\), \(\frac{26}{27}\), \(\frac{80}{81}\), and the \(n\)th partial sum is \(S_n = 2(1-(\frac{1}{3})^n)\).

Step 1: Identify the Sequence Term Formula

The sequence term formula is given by \(a_n = \frac{2}{3^n}\). This will be used to find the individual terms of the sequence.

Step 2: Calculate the First Four Terms

Calculate the first four terms of the sequence using the formula \(a_n = \frac{2}{3^n}\). For \(n = 1\), \(a_1 = \frac{2}{3^1} = \frac{2}{3}\) For \(n = 2\), \(a_2 = \frac{2}{3^2} = \frac{2}{9}\) For \(n = 3\), \(a_3 = \frac{2}{3^3} = \frac{2}{27}\) For \(n = 4\), \(a_4 = \frac{2}{3^4} = \frac{2}{81}\)

Step 3: Calculate Each Partial Sum

To find each partial sum, add the sequence terms up to that point.\[ S_1 = a_1 = \frac{2}{3} \] \[ S_2 = a_1 + a_2 = \frac{2}{3} + \frac{2}{9} = \frac{6}{9} + \frac{2}{9} = \frac{8}{9} \] \[ S_3 = a_1 + a_2 + a_3 = \frac{8}{9} + \frac{2}{27} = \frac{24}{27} + \frac{2}{27} = \frac{26}{27} \] \[ S_4 = a_1 + a_2 + a_3 + a_4 = \frac{26}{27} + \frac{2}{81} = \frac{78}{81} + \frac{2}{81} = \frac{80}{81} \]

Step 4: Derive the Formula for nth Partial Sum

Recognize that this sequence is geometric, where each term is multiplied by a common ratio of \(\frac{1}{3}\). Use the formula for the sum of the first \(n\) terms of a geometric sequence:\[ S_n = a \frac{1-r^n}{1-r} \] Here, \(a = \frac{2}{3}\), \(r = \frac{1}{3}\), thus:\[ S_n = \frac{2}{3} \frac{1-(\frac{1}{3})^n}{1-\frac{1}{3}} = 2(1-(\frac{1}{3})^n) \]

Step 5: Write the First Four Partial Sums and nth Partial Sum Formula

The first four partial sums are:\( S_1 = \frac{2}{3} \) \( S_2 = \frac{8}{9} \) \( S_3 = \frac{26}{27} \) \( S_4 = \frac{80}{81} \)The formula for the nth partial sum is:\( S_n = 2(1-(\frac{1}{3})^n) \).

Key Concepts

Partial Sums

A partial sum in a sequence is the sum of the first few terms. For example, consider the geometric sequence with terms given by the formula \(a_n = \frac{2}{3^n}\). Each term provides a distinct number, with the first term being \(\frac{2}{3}\), the second term \(\frac{2}{9}\), and so on. To find a partial sum, you sum these terms sequentially.

• The first partial sum \(S_1\) is simply the first term: \(S_1 = \frac{2}{3}\).
• The second partial sum \(S_2\) is the sum of the first two terms: \(S_2 = \frac{2}{3} + \frac{2}{9} = \frac{8}{9}\).
• The third partial sum \(S_3\) adds the first three terms: \(S_3 = \frac{2}{3} + \frac{2}{9} + \frac{2}{27} = \frac{26}{27}\).
• The fourth partial sum \(S_4\) accumulates the first four terms: \(S_4 = \frac{2}{3} + \frac{2}{9} + \frac{2}{27} + \frac{2}{81} = \frac{80}{81}\).

Partial sums are useful because they allow you to see how the sum of a sequence progresses as more terms are added.

Sequence Formula

The sequence formula is a tool used to find specific terms in a sequence. For this geometric sequence, the formula is \(a_n = \frac{2}{3^n}\). This defines each term based on its position \(n\) without needing to calculate the previous terms.

• The first term \(a_1\) is found by setting \(n = 1\), so \(a_1 = \frac{2}{3^1} = \frac{2}{3}\).
• For \(n = 2\), the term is \(a_2 = \frac{2}{9}\).
• For \(n = 3\), the term becomes \(a_3 = \frac{2}{27}\).
• Continuing, \(n = 4\) gives \(a_4 = \frac{2}{81}\).

Understanding the sequence formula is crucial because it provides a quick way to determine any term in the sequence directly, without recalculating all the preceding terms.

Geometric Series

A geometric series is the sum of the terms of a geometric sequence. Each term in the sequence is derived from the previous one by multiplying by a constant known as the common ratio, \(r\). For our sequence, the common ratio is \(\frac{1}{3}\).
To find the sum of the first \(n\) terms (i.e., the nth partial sum) of a geometric series, we use this formula: \[ S_n = a \frac{1-r^n}{1-r} \]where:

• \(a\) is the first term of the sequence, which is \(\frac{2}{3}\) in this case.
• \(r\) is the common ratio, here \(\frac{1}{3}\).

This formula lets us derive the partial sum without calculating each term individually. For example, substituting the values for our sequence:
\[ S_n = \frac{2}{3} \frac{1-(\frac{1}{3})^n}{1-\frac{1}{3}} = 2(1-(\frac{1}{3})^n) \]Geometric series are foundational in math, especially in calculations involving sums of rapidly decreasing sequences.