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Chapter 7: Problem 12

Find the sum of all the four-digit odd positive integers.

Short Answer

Expert verified
The sum of all the four-digit odd positive integers is \(24,750,000\).

Step by step solution

01

Find the range of four-digit odd positive integers

The smallest four-digit odd number is 1001, and the largest is 9999. These numbers are part of an arithmetic series with a common difference of 2, since each odd number can be obtained by adding two to the previous odd number.
02

Find the number of terms in the arithmetic series

To find the number of terms in this arithmetic series, we use the formula: \[n = \frac{L - a}{d} + 1,\] where 'n' represents the number of terms, 'L' is the last term (largest number), 'a' is the first term (smallest number), and 'd' is the common difference between the terms. In our case, this becomes: \[n = \frac{9999 - 1001}{2} + 1.\] Calculating the values, we get: \[n = \frac{8998}{2} + 1\] \[n = 4499 + 1\] \[n = 4500\] Therefore, there are 4500 terms in this series of four-digit odd numbers.
03

Find the sum of the four-digit odd positive integers

Now that we know there are 4500 terms in the series, we can use the arithmetic series formula for sum: \[S = \frac{n(a + L)}{2},\] where 'S' is the sum, 'n' is the number of terms, 'a' is the first term, and 'L' is the last term. Plugging in the values, we get: \[S = \frac{4500(1001 + 9999)}{2}.\] Simplifying and calculating the sum, we get: \[S = \frac{4500(11000)}{2}\] \[S = 4500 \times 5500\] \[S = 24750000\] Thus, the sum of all the four-digit odd positive integers is 24,750,000.

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Most popular questions from this chapter

Evaluate \(\lim _{n \rightarrow \infty} \frac{3 n+5}{2 n-7}\).

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