Question

Find the sum of the series. $$ 1 / 81+1 / 27+1 / 9+\cdots+243 $$

Short answer

Question: Find the sum of the given series: \(\frac{1}{81}, \frac{1}{27}, \frac{1}{9}, \frac{1}{3}, 243\) Answer: The sum of the given series is \(\frac{484}{81}\).

Step-by-step solution

Identify the first term, common ratio, and number of terms

In this series, the first term is \(\frac{1}{81}\), and we can see that each new term is obtained by multiplying the previous term by 3. Hence, the common ratio, r, is equal to 3. Now, to find the number of terms, we need to examine the last term given, which is 243. We can write it as a fraction: \(\frac{243}{1}\). If we divide this term by our first term, we will get the number of terms in the series. $$ \frac{\frac{243}{1}}{\frac{1}{81}} = \frac{243*81}{1} = 243*81 =3^5 $$ This tells us that there are 5 terms in the series.

Use the formula for the sum of a geometric series

The formula for the sum of a geometric series is given by: $$ S_n = \frac{a(1-r^n)}{1-r} $$ Where \(S_n\) is the sum of the series, \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms. In this problem, we have \(a = \frac{1}{81}\), \(r = 3\), and \(n = 5\). Plugging these values into the formula, we get: $$ S_5 = \frac{\frac{1}{81}(1-3^5)}{1-3} $$

Simplify and solve for the sum

Now we can simplify the expression and solve for \(S_5\): $$ S_5 = \frac{\frac{1}{81}(1-243)}{1-3} = \frac{\frac{1}{81}(-242)}{-2} $$ Multiplying the -242 by the -2 in the denominator, we get 484. Now, we can simplify the fraction: $$ S_5 = \frac{1}{81} * 484 = \frac{484}{81} $$ Thus, the sum of the given series is \(\frac{484}{81}\).

Key concepts

Common Ratio

In a geometric series, each term after the first is found by multiplying the previous term by a fixed, constant number. This constant is known as the "common ratio." It's one of the key elements that define a geometric progression.
For example:

• Start with a number, say 2.
• Multiply it by a common ratio, say 3, to get the next number: 2, 6, 18, and so on.

The concept of a common ratio is crucial because:

• It helps in determining the behavior of the sequence. A common ratio greater than 1 indicates an increasing series, while a ratio between 0 and 1 indicates a decreasing one.
• The common ratio also helps in defining the formula used to sum up all terms in the series, especially for finite sequences.

In our given problem, the common ratio is 3, indicating each term is three times the previous term.

Sum of Series

The sum of a geometric series refers to the total of all its terms. When the series is finite, there's a specific formula you can use to find this sum.
The formula is:\[ S_n = \frac{a(1-r^n)}{1-r} \]Where:

• \( S_n \) is the sum of the series.
• \( a \) is the first term of the series.
• \( r \) is the common ratio.
• \( n \) is the number of terms.

This formula is very useful for quickly finding the sum without having to manually add every term.
In our example, with \( a = \frac{1}{81} \), \( r = 3 \), and \( n = 5 \), substituting these values into the formula yields the sum \( \frac{484}{81} \).
The sum allows us to understand the overall "weight" or magnitude of a sequence, giving us insight into its growth or decrease over the given terms.

Number of Terms

Determining the number of terms in a geometric series is essential when applying the formula for the sum. Knowing how many times a number is multiplied by the common ratio provides insight into how far the series extends.
To find the number of terms:

• Identify the first term and divide it repeatedly by the common ratio until you reach the last term, or, as seen in the solution, calculate by dividing the last term \( (243) \) by the first term \( (\frac{1}{81}) \) and find how many times the ratio needs to be multiplied to reach from the first to the last term.

In the example provided, this method reveals there are 5 terms.
Knowing the number of terms is important for determining the end behavior of a sequence, allowing for precise calculations in problems involving finite series.