Solving a mathematical problem about proposition and proof The following is the definition of "mysterious number": a positive integer that can be expressed as the square difference of two consecutive even numbers is called a mysterious number. Please judge 4, 12, 20, 28, 2012 as a mysterious number according to the definition of "mysterious number". Why?

Solving a mathematical problem about proposition and proof The following is the definition of "mysterious number": a positive integer that can be expressed as the square difference of two consecutive even numbers is called a mysterious number. Please judge 4, 12, 20, 28, 2012 as a mysterious number according to the definition of "mysterious number". Why?


(2n-2) ^ 2 - (2n) ^ 2 = 8N + 4 = 4 (2n + 1) is a mysterious number, which is an odd multiple of 4
So 12, 28, 2012 are mysterious numbers



(1 / 2 + 2 / 3 + 3 / 4 + 4 / 5 + 5 / 6 + 6 / 7) + (1 / 2 + 2 / 3 + 3 / 4 + 4 / 5 + 5 / 6 + 6 / 7) * 1 / 2 - (1 + 1 / 2 + 2 / 3 + 3 / 4 + 4 / 5 + 5 / 6 + 6 / 7) * (2 / 3 + 3 / 4 + 4 / 5 + 5 / 6 + 6 / 7)


(1/2+2/3+3/4+4/5+5/6+6/7)+(1/2+2/3+3/4+4/5+5/6+6/7)×1/2-(1+1/2+2/3+3/4+4/5+5/6+6/7)×(2/3+3/4+4/5+5/6+6/7)
=(1/2+2/3+3/4+4/5+5/6+6/7)×(1+1/2)-[(1+1/2)+(2/3+3/4+4/5+5/6+6/)]×(2/3+3/4+4/5+5/6+6/7)
=(3/2)×(1/2+2/3+3/4+4/5+5/6+6/7)-(3/2)×(2/3+3/4+4/5+5/6+6/7)-(2/3+3/4+4/5+5/6+6/7)×(2/3+3/4+4/5+5/6+6/7)
=(3/2)×[(1/2+2/3+3/4+4/5+5/6+6/7)-(2/3+3/4+4/5+5/6+6/)]-(2/3+3/4+4/5+5/6+6/7)^2
=(3/2)×(1/2)-(2/3+3/4+4/5+5/6+6/7)^2
=3/4-(2/3+3/4+4/5+5/6+6/7)^2
2/3+3/4+4/5+5/6+6/7
=(1-1/3)+(1-1/4)+(1-1/5)+(1-1/6)+(1-1/7)
=5-(1/3+1/4+1/5+1/6+1/7)
The least common multiple of 3,4,5,6,7 is 4 × 3 × 5 × 7 = 420
1/3+1/4+1/5+1/6+1/7
=(4×5×7+3×5×7+4×3×7+2×5×7+4×3×5)/420
=(140+105+84+70+60)/420
=459/420
=153/140
2/3+3/4+4/5+5/6+6/7=5-153/140=547/140
Original formula = 3 / 4 - (547 / 140) ^ 2 = - 284509 / 19600 ≈ - 14.51576531



Let u = {positive integer no more than 5}, a = {x | x2-5x + q = 0}, B = {x | x2 + PX + 12 = 0}, (∁ UA) ∪ B = {1, 3, 4, 5}, find P, Q and set a, B


The complete set u = {1,2,3,4,5}, a = {x | x2-5x + q = 0}, B = {x | x2 + PX + 12 = 0}, (∁ UA) ∪ B = {1,3,4,5}, ∪ 2 ∈ a, substituting x = 2 into x2-5x + q = 0, we get: 4-10 + q = 0, namely q = 6, namely x2-5x + 6 = 0, ∁ X-2) (x-3) = 0, namely x = 2 or x = 3, ∁ a = {2,3}, ∁ UA = {1,4