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Chapter 7: Problem 2

Evaluate the arithmetic series. $$ 1001+1002+1003+\cdots+2998+2999+3000 $$

Short Answer

Expert verified
The sum of the arithmetic series \(1001+1002+\cdots+3000\) can be found using the formula \(S_n = \frac{n}{2}(a_1+a_n)\). First, identify the first term (\(a_1=1001\)) and the last term (\(a_n=3000\)). Next, find the number of terms by using the formula \(n = \frac{(a_n - a_1)}{d} + 1\), which gives \(n=2000\). Finally, plug the values into the sum formula to get \(S_n = \frac{2000}{2}(1001+3000) = 4,001,000\).

Step by step solution

01

Identifying the first and last terms

The first term of the series is \(a_1 = 1001\), and the last term is \(a_n = 3000\).
02

Finding the common difference

The common difference in an arithmetic series is the difference between consecutive terms. In this series, the common difference is \(1002 - 1001 = 1\).
03

Calculating the number of terms

To find the number of terms in the series, we can use the formula: $$ n = \frac{(a_n - a_1)}{d} + 1, $$ where \(d\) is the common difference. Plugging in the values, we get: $$ n = \frac{(3000 - 1001)}{1} + 1 = 2000. $$
04

Evaluating the arithmetic series

Now, we can use the sum formula: $$ S_n = \frac{n}{2}(a_1+a_n) $$ Plugging in the values, we get: $$ S_n = \frac{2000}{2}(1001+3000) = 1000(4001) = 4,001,000. $$ So, the sum of the arithmetic series is 4,001,000.

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