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Find the first n terms and S N of sequence 1 * 2 / 2,2 * 3 / 2,3 * 4 / 2,4 * 5 / 2
Sn = 1 × 2 / 2 + 2 × 3 / 2 + 3 × 4 / 2 + 4 × 5 / 2 + (n-1) 2 / N = 2 × [1 × 2 / 1 + 2 × 3 / 1 + 3 × 4 / 1 + 4 × 5 / 1 + +(n-1) 1 / N] = 2 × [1-2 / 1 + 2 / 1-3 / 1 + 3 / 1-4 / 1 + 4 / 1-5 / 1 + +(n-1) 1 / 2 -
Find the sum of the first n terms of the sequence 1 / 1 * 2 * 3,1 / 2 * 3 * 4,1 / 3 * 4 * 5,1 / 4 * 5 * 6
1 / 3 * 4 * 5 = 1 / 3 * 4-1 / 4 * 5.
The upper two digits together are positive solutions 1 / 1 * 2 * 3 + 1 / 2 * 3 * 4 + 1 / 3 * 4 * 5 + 1 / 4 * 5 * 6 = {(1 / 1 * 2-1 / 2 * 3) + (1 / 2 * 3-1 / 3 * 4) +. + [1 / N * (n + 1) - 1 / (n + 1) * (n + 2)]} / 2 = [1 / 2-1 / (n + 1) * (n + 2)] / 2 = [(n + 1) (n + 2) - 2] / 4 (n + 1) (n + 2)
Fill in the numbers according to the rules______ 、______ ．
Because 40-16 = 24, 24-8 = 16, 16-4 = 12, the subtractions are 16, 8, 4 respectively, there is such a relationship between the subtractions; 16 △ 8 = 4, then the next subtraction is 8 △ 4 = 2, this number is: 12-2 = 10; the next subtraction is: 2 △ 2 = 1, this number is: 10-1 = 9. So the answer is: 10, 9
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