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Chapter 9: Problem 82

Use a graphing utility to find the partial sum. $$\sum_{j=1}^{200}(10.5+0.025 j)$$

Short Answer

Expert verified
The partial sum of the series is 2597.5.

Step by step solution

01

Identify the Arithmetic Series Terms

Identify the arithmetic series components in the exercise. The series starts from the first term 10.5 (which is \(a_1\)), increments by 0.025 (the common difference, \(d\)), and ends at the 200th term (the number of terms, \(n\)). Thus, \(a_1 = 10.5\), \(d = 0.025\) and \(n = 200\).
02

Apply the Arithmetic Series Formula

The sum of the arithmetic series is found by applying the formula: \(S = n/2 * [2a + (n - 1)d]\). Substituting the identified components into this formula gives: \(S = 200/2 * [2 * 10.5 + (200 - 1) * 0.025]\). This simplifies to \(S = 100 * [21 + 199 * 0.025]\).
03

Calculate the Sum

Simplify the expression by carrying out the arithmetic operations: 199*0.025 equals to 4.975, resulting the equation \(S = 100 * [21 + 4.975]\). Further simplify this equation: \(S = 100 * 25.975\). Lastly, calculate the sum \(S = 2597.5\).

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