Problem 64

Question

A partial sum of an arithmetic sequence is given. Find the sum. $$\sum_{n=0}^{20}(1-2 n)$$

Step-by-Step Solution

Verified

Answer

The sum is -399.

1Step 1: Identify the Arithmetic Sequence

Given the sum $ \sum_{n=0}^{20}(1-2n) $, we need to identify the arithmetic sequence. For each term $ a_n = 1 - 2n $, the first term $ a_0 $ is 1 when $ n = 0 $, the second term $ a_1 $ is -1 when $ n = 1 $, and the common difference $ d $ is -2 (since $ a_1 - a_0 = -1 - 1 = -2 $).

2Step 2: Determine the Number of Terms

The sequence starts at $ n = 0 $ and goes up to $ n = 20 $. Thus, the number of terms $ N $ is $ 20 - 0 + 1 = 21 $ terms.

3Step 3: Find the Last Term of the Sequence

The last term $ a_{20} $ is obtained by substituting $ n = 20 $ in the formula $ a_n = 1 - 2n $. So, $ a_{20} = 1 - 2 \times 20 = 1 - 40 = -39 $.

4Step 4: Calculate the Sum of the Arithmetic Sequence

The formula for the sum of an arithmetic sequence is $ S_N = \frac{N}{2} \times (a_0 + a_{N-1}) $. Substituting the known values: $ N = 21 $, $ a_0 = 1 $, and $ a_{N-1} = a_{20} = -39 $, we get $ S_{21} = \frac{21}{2} \times (1 - 39) = \frac{21}{2} \times (-38) = 21 \times (-19) = -399 $.

5Step 5: Verify Your Solution

Double-check your calculations to ensure consistency. Confirm that arithmetic sequence properties and the steps have been followed correctly. The sum $ S_{21} = -399 $ should be verified for consistency with calculations and formula applications.

Key Concepts

Arithmetic SeriesPartial SumCommon DifferenceSequence Sum Formula

Arithmetic Series

An arithmetic series is simply the sum of the terms in an arithmetic sequence. An arithmetic sequence has a constant difference between consecutive terms, known as the common difference. The series involves adding up these terms. In the exercise above, we have an arithmetic sequence defined by each term as $ a_n = 1 - 2n $, where $ n $ ranges from 0 to 20.
The series involves summing these calculated terms from $ a_0 $ to $ a_{20} $. Understanding the structure of the arithmetic sequence, such as finding the first term and the common difference, is crucial to formulating and solving an arithmetic series.

Partial Sum

The term partial sum refers to summing only a part of an infinite sequence. Specifically, it covers only a certain number of initial terms of a sequence. The sum calculated as part of the question is a partial sum because it only sums terms from $ n = 0 $ up to $ n = 20 $, rather than an infinite number of terms.
The partial sum in arithmetic sequences can be found using specific formulas, as was done in the exercise. Here, we use a formula specifically for finite arithmetic sequences, which ensures that we cover the exact range needed for our problem of 21 terms.

Common Difference

The common difference in an arithmetic sequence is the constant difference between any two successive terms. Recognizing this pattern is important, as it helps in calculating terms throughout the sequence.

In our specific exercise, the common difference $ d $ is determined by finding the difference between the first two terms: $ a_1 - a_0 = -1 - 1 = -2 $.
This consistent step of subtracting 2 continues throughout the sequence, helping to find any term $ a_n $ when needed.

Understanding the common difference is key because it aids in constructing each subsequent term in the sequence and leads to the potential sum formulation.

Sequence Sum Formula

The sequence sum formula is a powerful tool for finding the sum of arithmetic sequences. This formula requires understanding both the number of terms and the values of the first and last terms. The formula to find the sum $ S_N $ of the first $ N $ terms in an arithmetic sequence is:
\[ S_N = \frac{N}{2} \times (a_0 + a_{N-1}) \]

In our scenario, $ N = 21 $, $ a_0 = 1 $, and $ a_{N-1} = -39 $.
By plugging these into the formula, the sum $ S_{21} = \frac{21}{2} \times (1 - 39) $.
The computation ultimately yields $ S_{21} = -399 $.

This formula provides an efficient way to calculate sums without manually adding up all the individual terms, saving time and preventing errors.

Problem 63

Problem 64

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Problem 63

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Problem 64

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Problem 64

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