Problem 64 - Precalculus: Mathematics for Calculus

Question

Find the sum. $$\sum_{k=1}^{6} 64\left(\frac{3}{2}\right)^{k-1}$$

Step-by-Step Solution

Answer

The sum is 1330.

Step 1: Identify the Type of Series

The sum given is an example of a geometric series. In general, a geometric series has the form \( a + ar + ar^2 + \ldots + ar^{n-1} \), where \( a \) is the first term and \( r \) is the common ratio.

Step 2: Determine the First Term and Common Ratio

For the series \( 64\left(\frac{3}{2}\right)^{k-1} \), the first term \( a \) is \( 64 \), and the common ratio \( r \) is \( \frac{3}{2} \).

Step 3: Use the Sum Formula for a Geometric Series

The sum of the first \( n \) terms of a geometric series is given by the formula:\[S_n = a \frac{r^n - 1}{r - 1}\]In this problem, \( n = 6 \), \( a = 64 \), and \( r = \frac{3}{2} \).

Step 4: Plug the Values into the Formula

Using the formula:\[S_6 = 64 \frac{\left(\frac{3}{2}\right)^6 - 1}{\frac{3}{2} - 1}\]First, calculate the numerator \( \left(\frac{3}{2}\right)^6 - 1 \).

Step 5: Calculate the Numerical Parts

Calculate \( \left(\frac{3}{2}\right)^6 \): \[(\frac{3}{2})^6 = \frac{729}{64}\]Therefore, \( \frac{729}{64} - 1 = \frac{729 - 64}{64} = \frac{665}{64} \).

Step 6: Simplify the Denominator

The denominator \( \frac{3}{2} - 1 = \frac{1}{2} \).

Step 7: Compute the Sum

Now compute:\[S_6 = 64 \times \frac{\frac{665}{64}}{\frac{1}{2}} = 64 \times \frac{665}{64} \times 2 = 1330\]The sum of the series is 1330.

Key Concepts

First Term

In a geometric series, the **first term** is pivotal. It sets the initial value from which all subsequent terms are derived. Represented often as "\( a \)," the first term is the starting point of the series. For the series in the exercise, the first term \( a \) is given as 64. Understanding this concept is important because each term after the first is calculated by multiplying the previous term by a constant known as the common ratio.

Knowing \( a \) helps with:

• Determining subsequent terms in the series
• Calculating the sum of the series

Thus, the first term is essential in both constructing the series and computing any sums.

Common Ratio

The **common ratio** in a geometric series is the factor by which we multiply each term to get the next term. In this discussion, the symbol "\( r \)" denotes it. It shows how each term in the series changes relative to the one before it. For our example, the common ratio \( r \) is \( \frac{3}{2} \).

This constant ratio is central to identifying the nature of the series:

• If \( r > 1 \), the terms increase.
• If \( 0 < r < 1 \), the terms decrease.
• If \( r = 1 \), the series is trivial (all terms are equal).
• If \( r < 0 \), the sequence alternates in sign.

To compute the series, knowing \( r \) helps determine the pattern and progression of the terms.

Sum Formula for Geometric Series

To find the sum of the first *n* terms in a geometric series, we use a formula that takes into account both the first term and the common ratio. The **sum formula for a geometric series** is:\[S_n = a \frac{r^n - 1}{r - 1}\]Here, \( S_n \) represents the sum of the first \( n \) terms, \( a \) is the first term, and \( r \) is the common ratio. This formula helps simplify the process of adding up all the terms without having to perform each multiplication individually.

For example, to find the sum of the first 6 terms in our series:

• Plug \( a = 64 \), \( r = \frac{3}{2} \), and \( n = 6 \) into the formula.
• Simplify the expression by calculating \( (\frac{3}{2})^6 \) and subtract 1.
• Divide by \( r - 1 \), which simplifies to \( \frac{1}{2} \).
• Finally, multiply by \( a \) to get the total sum.

The result is the total sum of the series is 1330, showcasing the efficiency of this formula in finding sums of geometric progressions.