Given the set a = {2,3, A2 + 1}, B = {A2 + a-4,2a + 1,1}, and a ∩ B = {2}, find the value of A

Given the set a = {2,3, A2 + 1}, B = {A2 + a-4,2a + 1,1}, and a ∩ B = {2}, find the value of A

A ∩ B = {2}, so A2 + A-4 or 2A + 1 = 2
The solution is a = - 3 or 2, or 1 / 2
Test: a = - 3, true
A = 2, A2 + 1 = 2A + 1 = 5
A = 1 / 2, yes
So a = - 3 or 1 / 2

As shown in the figure, Al (1,0), A2 (1,1), A3 (- 1,1), A4 (- 1, - 1), A5 (2, - 1) . then the coordinates of point a2007 are______ ．

It can be seen from the figure and narration that all points (except A1 point and the point in the fourth quadrant) are located on the angle bisector of the quadrant. The subscripts of the letters corresponding to the point in the first quadrant are 2, 6, 10, 14, that is 4n-2 (n is the natural number, n is the absolute value of the abscissa of the point); similarly, the subscripts of the points in the second quadrant are 4N-1 (n is the natural number, n is the absolute value of the abscissa of the point) The third quadrant is 4N (n is the natural number, n is the absolute value of the abscissa of the point); the fourth quadrant is 1 + 4N (n is the natural number, n is the absolute value of the abscissa of the point); 2007 = 4N-1, then n = 502, when 2007 is equal to 4N + 1 or 4N or 4n-2, there is no such value of N. therefore, the point a2007 is on the angular bisector of the second quadrant, that is, the coordinate is (- 502502)

(a1+a2…… +a2008)（a2+a3+…… a2009)-(a2+a3+…… +a2008)(a1+a2+…… a2009)

Let m = a1 + A2 + +A2008 is A2 + a3 + a2009=m-a1+a2009a2+a3+…… +a2008=m-a1a1+a2+…… A2009 = m + A2009, so the original formula = m (m-a1 + A2009) - (m-a1) (M + A2009) = m ^ 2-m (a1-a2009) - m ^ 2 + m (a1-a2009) + A1 * A2009 = A1 * A2009