Problem 67
Question
Use a graphing calculator to evaluate each sum. Round to the nearest thousandth. $$\sum_{j=3}^{8} 2(0.4)^{j}$$
Step-by-Step Solution
Verified Answer
0.212
1Step 1: Understand the Sum Notation
The given sum is a series where we need to calculate the sum of the function \( 2(0.4)^j \) as \( j \) goes from 3 to 8. This means we will calculate the term for each value of \( j \) from 3 to 8 and then find the total sum of these terms.
2Step 2: Compute Each Term
Use the expression \( 2(0.4)^j \) to calculate each term of the series. Substitute \( j = 3 \), \( j = 4 \), \( j = 5 \), \( j = 6 \), \( j = 7 \), and \( j = 8 \) respectively:1. For \( j = 3 \): \( 2(0.4)^3 = 2(0.064) = 0.128 \)2. For \( j = 4 \): \( 2(0.4)^4 = 2(0.0256) = 0.0512 \)3. For \( j = 5 \): \( 2(0.4)^5 = 2(0.01024) = 0.02048 \)4. For \( j = 6 \): \( 2(0.4)^6 = 2(0.004096) = 0.008192 \)5. For \( j = 7 \): \( 2(0.4)^7 = 2(0.0016384) = 0.0032768 \)6. For \( j = 8 \): \( 2(0.4)^8 = 2(0.00065536) = 0.00131072 \)
3Step 3: Sum the Terms
Add together all the computed terms from the previous step: \[ 0.128 + 0.0512 + 0.02048 + 0.008192 + 0.0032768 + 0.00131072 \]Perform the addition to find the total sum which equals \( 0.21245952 \).
4Step 4: Round the Sum
Round the calculated sum (\( 0.21245952 \)) to the nearest thousandth. The third decimal digit is 2, so the sum rounds to \( 0.212 \).
Key Concepts
Series EvaluationSum NotationRounding Decimals
Series Evaluation
When we talk about series evaluation, we mean calculating the total sum of numbers that are part of a sequence. A series is essentially a sum that runs over a series of terms, following some pattern or rule. In our exercise, we are evaluating the series of the function \( 2(0.4)^j \) where \( j \) ranges from 3 to 8.
To understand this better, think of each function value as a step in a ladder. To find the total height (or the sum of the series), you add each step from the third step to the eighth step. Using a graphing calculator can simplify this process as it quickly computes each term, and subsequently adds them together.
In practice, this means inputting the function into your calculator repeatedly for \( j = 3, 4, 5, 6, 7, \) and \( 8 \). Once each term is determined, you add them all together to arrive at the full series sum.
To understand this better, think of each function value as a step in a ladder. To find the total height (or the sum of the series), you add each step from the third step to the eighth step. Using a graphing calculator can simplify this process as it quickly computes each term, and subsequently adds them together.
In practice, this means inputting the function into your calculator repeatedly for \( j = 3, 4, 5, 6, 7, \) and \( 8 \). Once each term is determined, you add them all together to arrive at the full series sum.
Sum Notation
Sum notation, often represented with the Greek letter Sigma (\( \Sigma \)), is a concise way to express the addition of a series of terms. Just like a recipe tells you the ingredients needed for a dish, sum notation lays out all you need to know to compute a series.
Our problem uses the notation \( \sum_{j=3}^{8} 2(0.4)^{j} \), which tells you to substitute each integer from 3 to 8 for \( j \) in the formula \( 2(0.4)^j \) and then add them together. It's a systematic approach to calculating more complex series, saving you from having to write out each term manually.
Our problem uses the notation \( \sum_{j=3}^{8} 2(0.4)^{j} \), which tells you to substitute each integer from 3 to 8 for \( j \) in the formula \( 2(0.4)^j \) and then add them together. It's a systematic approach to calculating more complex series, saving you from having to write out each term manually.
- The lower number, \( 3 \), is the starting point for our calculations, indicating where our sum begins.
- The upper number, \( 8 \), tells us where the calculation ends.
- The expression \( 2(0.4)^j \) is what you plug each of those numbers into, to find individual terms.
Rounding Decimals
Rounding decimals is the process of adjusting a number to make it simpler and shorter, while keeping its value close to what it was initially. This is particularly useful in mathematical problems where precision is important, but an approximate result can suffice.
In this exercise, we had a calculated sum of \( 0.21245952 \). The instruction was to round this to the nearest thousandth. Observing the number, the importance lies in the digits beyond the thousandth place:
In this exercise, we had a calculated sum of \( 0.21245952 \). The instruction was to round this to the nearest thousandth. Observing the number, the importance lies in the digits beyond the thousandth place:
- The digit in the thousandth place was 2, which is retained as is.
- The digit immediately following it was 4. Since this is less than 5, the thousandth place remains unchanged.
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