Problem 20 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Express each of the sums in closed form. Wherever possible, give a numerical approximation of the sum, rounded off to 3 decimal places. $$ \frac{2}{3}+2+6+\cdots+2(3)^{100} $$

Step-by-Step Solution

Answer

The closed form of the given series is \(-\frac{1}{3}(1 - 3^{101})\)

Step 1: Recognize the series

The term \(2(3)^{0}\) equals 2, \(2(3)^{1}\) equals 6, and \(2(3)^{2}\) equals 18, and so on. We observe that the given series:\[\frac{2}{3}+2+6+...+2(3)^{100}\]is indeed a geometric series with first term \(a = \frac{2}{3}\) and the common ratio \(r = 3\).

Step 2: Apply the formula for the sum of a geometric series

The sum \(S_n\) of the first n terms of a geometric series where the first term is 'a' and the common ratio is 'r' is given by:\[S_n = \frac{a \cdot (1 - r^{n+1})}{1 - r}, n \geq 0\]Now, apply this formula by replacing 'a' with \(\frac{2}{3}\), 'r' with 3, and 'n' with 100 to find the closed form of the series.

Step 3: Calculate the sum

Substituting the values we get:\[S_{100} = \frac{\frac{2}{3} \cdot (1 - 3^{101})}{1 - 3} = -\frac{1}{3}(1 - 3^{101})\]Then simplify this expression to find the numerical approximation of the sum.

Step 4: Numerical Approximation

To find the numerical approximation, we calculate the value of the expression \(-\frac{1}{3}(1 - 3^{101})\) using a calculator and round off to 3 decimal places.

Key Concepts

Closed Form Expression

The term "closed form expression" describes a way to represent an infinite or finite series as a single, compact formula. This approach can greatly simplify complicated series and make computations easier. In our exercise, we're dealing with a geometric series, which is a series where each term is a consistent multiple of the previous one.

For a geometric series, such as the one given in the exercise, the closed form expression can be determined using the formula:

• First term (\(a\)): \(\frac{2}{3}\)
• Common ratio (\(r\)): 3
• Number of terms (\(n\)): 100

The closed form of the sum of the first \(n\) terms is given by the formula:\[S_n = \frac{a \cdot (1 - r^{n+1})}{1 - r}\] Inserting the provided values, we eventually arrive at the expression:\[S_{100} = -\frac{1}{3}(1 - 3^{101})\]

This transformation allows us to condense a potentially bafflingly large amount of terms into a simple formula. Understanding it helps in both theoretical and numerical practices.

Numerical Approximation

Once we have derived the closed form expression, the next step often involves finding a numerical approximation. This can particularly help us understand the magnitude of the sum without manually adding all the terms.

In our example, we derive:\[S_{100} = -\frac{1}{3}(1 - 3^{101})\]Calculating this directly can be complex due to the enormous size of \(3^{101}\). Therefore, numerical approximation involves using calculators or computational tools to compute the value and then simplifying it to a more comprehensible form, often rounding the result to a certain number of decimal places for precision.

• Use a scientific calculator or software to compute \(-\frac{1}{3}(1 - 3^{101})\).
• Round the answer to 3 decimal places for a clearer representation.

This process not only provides us with a numerical insight into the closed form but also reinforces the practical ease of handling large calculations and approximations efficiently.

Sum of Series

The sum of a series is essentially the combined total of all its terms. In the context of geometric series, finding this sum involves a specific method due to the multiplicative pattern of its terms.

Consider the geometric series from the exercise: \[\frac{2}{3} + 2 + 6 + \cdots + 2(3)^{100}\]To find the sum, we use the principle formula for geometric series:\[S_n = \frac{a \cdot (1 - r^{n+1})}{1 - r}\]This allows for the addition of terms over a potentially huge range, even for very high \(n\) values, without manual addition.

Using the previously obtained closed form expression effectively shifts the task of summing an entire series into a manageable calculation using merely multiplication and division. This underscores the elegance and utility of mathematical formulas in real-world applications, such as finance and physics, where handling geometric growth models is a regular necessity.