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What is the sum of 1 times 2 plus 3 times 4 plus 5 times 6 plus 7 times 8 plus 9 times 10 plus. 99 times 100? It's a simple operation
Let the original formula = m be (1) 1 ^ 2 + 1 + 3 ^ 2 + 3 + 5 ^ 2 + 5 +. + 99 ^ 2 + 99 (2) 2 ^ 2-2 + 4 ^ 2-4 + 6 ^ 2-6 +. + 100 ^ 2-100 (1) + (2) = 2m. Use the formula 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + +n^2=n(n+1)(2n+1)/62M=[100(100+1)(200+1)/6]-50 = 33830M=169150
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- 19. 1 + 2 + 1 = 4,1 + 2 + 3 + 2 + 1 = 9,1 + 2 + 3 + 4 + 3 + 2 + 1 = 16, calculate (1) 1 + 2 + 3 + +99+100＝?
- 20. Observe the following formulas: 1 + 2 + 1 = 4, 1 + 2 + 3 + 2 + 1 = 9, 1 + 2 + 3 + 4 + 3 + 2 + 1 = 16, 1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 25 According to the rule you found, please write the result of the following formula directly: 1 + 2 + 3 + +99+100+99+… +3+2+1=______ ．