Chapter 5 | Integration
511
b. The sum of the terms ⎛
⎞ ⎠ for i =1,2,3,4,5,6.
⎝ i 3 − i 2
Solution a. Multiplying out ( i −3) 2 , we can break the expression into three terms. ∑ i =1 200 ( i −3) 2 = ∑ i =1 200 ⎛ ⎝ i 2 −6 i +9 ⎞ ⎠
= ∑ i =1 200 = ∑ i =1 200
i 2 − ∑ i =1 200 i 2 −6 ∑ i =1 200
6 i + ∑ i =1 200 i + ∑ i =1 200
9
9
⎡ ⎣ 200(200 + 1) 2 ⎤
⎦ +9(200)
= 200(200 + 1)(400 + 1) 6
−6
= 2,686,700 − 120,600 + 1800 = 2,567,900 b. Use sigma notation property iv. and the rules for the sum of squared terms and the sum of cubed terms. ∑ i =1 6 ⎛ ⎝ i 3 − i 2 ⎞ ⎠ = ∑ i =1 6 i 3 − ∑ i =1 6 i 2
2 (6+1) 2
6(6+1) ⎛
⎝ 2(6)+1 ⎞ ⎠ 6
= 6
4 −
546 6
= 1764 4 −
=350
Find the sum of the values of 4+3 i for i = 1, 2,…, 100.
5.2
Example 5.3 Finding the Sum of the Function Values
Find the sum of the values of f ( x ) = x 3 over the integers 1, 2, 3,…, 10.
Solution Using the formula, we have
∑ i =0 10
2 (10+1) 2 4
i 3 = (10)
= 100(121) 4 =3025.
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