What is the sum of the series?
[tex]\sum_{j=1}^{38}\left(j^3-25 j\right)[/tex]
Answer (2)
We are asked to find the sum of the series ∑ j = 1 38 ( j 3 − 25 j ) .
Split the series into two separate sums: ∑ j = 1 38 j 3 and ∑ j = 1 38 25 j .
Calculate ∑ j = 1 38 j 3 = ( 2 38 ( 38 + 1 ) ) 2 = 549081 .
Calculate ∑ j = 1 38 25 j = 25 ∑ j = 1 38 j = 25 ⋅ 2 38 ( 38 + 1 ) = 18525 .
Subtract the second sum from the first sum: 549081 − 18525 = 530556 .
The sum of the series is 530556 .
Explanation
Understanding the Problem We are asked to find the sum of the series ∑ j = 1 38 ( j 3 − 25 j ) . This means we need to add up the values of the expression j 3 − 25 j for each integer j from 1 to 38.
Breaking Down the Series We can split the series into two separate sums: ∑ j = 1 38 j 3 and ∑ j = 1 38 25 j . This allows us to use known formulas for the sum of cubes and the sum of integers.
Calculating the Sum of Cubes The formula for the sum of the first n cubes is ∑ j = 1 n j 3 = ( 2 n ( n + 1 ) ) 2 . Plugging in n = 38 , we get j = 1 ∑ 38 j 3 = ( 2 38 ( 38 + 1 ) ) 2 = ( 2 38 ⋅ 39 ) 2 = ( 19 ⋅ 39 ) 2 = 74 1 2 = 549081.
Calculating the Sum of Integers The formula for the sum of the first n integers is ∑ j = 1 n j = 2 n ( n + 1 ) . Therefore, j = 1 ∑ 38 25 j = 25 j = 1 ∑ 38 j = 25 ⋅ 2 38 ( 38 + 1 ) = 25 ⋅ 2 38 ⋅ 39 = 25 ⋅ ( 19 ⋅ 39 ) = 25 ⋅ 741 = 18525.
Finding the Total Sum Now, we subtract the second sum from the first sum: 549081 − 18525 = 530556.
Final Answer Therefore, the sum of the series is 530556.
Examples
Imagine you are stacking boxes in a warehouse. The number of boxes on each level follows a cubic pattern (1 box on the first level, 2 3 = 8 boxes on the second level, 3 3 = 27 on the third, and so on). However, you need to subtract a certain number of boxes from each level due to space constraints, specifically 25 times the level number. The sum of the series helps you calculate the total number of boxes you'll have after stacking 38 levels, considering the cubic increase and the linear reduction. This type of calculation can be useful in inventory management, logistics, and resource planning.
The sum of the series ∑ j = 1 38 ( j 3 − 25 j ) is calculated by finding the difference between the sum of cubes and a linear sum. The total result is 530556 .
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