Problem 9 - Calculus Early Transcendentals

Question

Evaluate each geometric sum. $$\sum_{k=0}^{20}\left(\frac{2}{5}\right)^{2 k}$$

Step-by-Step Solution

Answer

Question: Evaluate the geometric sum \(\displaystyle\sum_{k=0}^{20}\left(\frac{2}{5}\right)^{2 k}\). Answer: The sum of the given geometric series is approximately \(8.943\).

Step 1: Identify the sequence and common ratio

The given geometric sum is \(\displaystyle\sum_{k=0}^{20}\left(\frac{2}{5}\right)^{2 k}\). We notice that the common ratio of the series is \(r = \left(\frac{2}{5}\right)^2\) as the exponent of the term is \(2k\).

Step 2: Use the geometric series sum formula

The formula for the sum of a geometric series is given by \(S_n = \frac{a_1(1-r^n)}{1-r}\), where \(n\) is the number of terms, \(a_1\) is the first term, and \(r\) is the common ratio.

Step 3: Substitute the values into the sum formula

In our case, the first term \((a_1)\) is obtained when \(k = 0\), so \(a_1 = \left(\frac{2}{5}\right)^{0} = 1\). There are \(n= 21\) terms in the sequence as the series goes from \(k=0\) to \(k=20\), so we will substitute these values and the common ratio into the sum formula: \(S_{21} = \frac{1(1-\left(\frac{2}{5}\right)^{2 \cdot 20})}{1-\left(\frac{2}{5}\right)^2}\).

Step 4: Simplify the expression

Now, we will simplify the expression to get the sum of the series: \(S_{21} = \frac{1(1-\left(\frac{2}{5}\right)^{40})}{1-\left(\frac{2}{5}\right)^2}\).

Step 5: Evaluate the sum

After calculating the sum, we get: \(S_{21} = \frac{1(1-\left(\frac{2}{5}\right)^{40})}{\frac{21}{25}}\). Then by dividing each term in the numerator by the denominator, we obtain: \(S_{21} = \frac{1 - \left(\frac{2}{5}\right)^{40}}{\frac{21}{25}} = \frac{25}{21}(1- \left(\frac{2}{5}\right)^{40}) \approx 8.943\). So, the sum of the given geometric series is approximately \(8.943\).