Fill in the eight vertices of a square with the eight numbers of - 1, - 2, - 3, - 4, - 5, - 6, - 7, - 8. Each vertex can only fill in one number, so that the sum of the four vertices on the six faces of the square is equal
The square is - 18
Add - 8, - 3, - 6, - 1 clockwise from the left on the front and - 2. - 5, - 4, - 7 clockwise from the left on the back,
The sum of each side is - 18
Syndrome 2 / 1 * 4 / 3 * 6 / 5 * 8 / 7 *. (2n) / (2n-1) > √ (2n + 1)
2^2 > 1*3
4^2 > 3*5
.
(2n)^2 > (2n-1)(2n+1)
∴(2*4*6*8*...2n)^2 > 1*3^2*5^2*7^2*...(2n-1)^2*(2n+1)
∴[(2*4*6*8*...2n)/(1*3*5*7*...(2n-1))]^2 > 2n+1
∴2/1*4/3*6/5*8/7*.(2n)/(2n-1) > √(2n+1)
What is 1 / (1 × 3) + 1 / (3 × 5) + 1 / (5 × 7) +. + 1 / (2n-1) (2n + 1) equal to?
=1/2(1-1/3+1/3-1/5+…… +1/(2n-1)-1/(2n+1))
1/2(1-1/(2n+1))=n/2n+1