Known sequence {an}, {BN} satisfy: A1 = 1 / 4, an + BN = 1, BN + 1 = BN / 1-an ^ 2. Find {BN} general term formula

A (n + 1) + B (n + 1) = 1, B (n + 1) = (1-an) / (1-an & # 178;) = 1 / (1 + an), a (n + 1) + 1 / (1 + an) = 1, a (n + 1) an + a (n + 1) + 1 = 1 + an, a (n + 1) an + a (n + 1) = an, 1 / a (n + 1) - 1 / an = 1, the sequence {1 / an} is an arithmetic sequence, the tolerance is 1, the first term is 4, 1 / an = 4 + n-1 = n + 3, an = 1 / (n + 3), B (n + 1) = 1 /

The image of the quadratic function y = x2-8x + 15 intersects with the x-axis at L and m, and the point n moves on the image of the function, so that the point n with the area of △ LMN equal to 2 shares ()
A. 0 B. 1 C. 2 d. 3

Let y = 0, we get x2-8x + 15 = 0, the solution is X1 = 3, X2 = 5, ∧ L (3,0), m (5,0) LM = 5-3 = 2, the area of ∧ LMN is equal to 2, the ordinate of ∧ n point is 2 or - 2, when y = 2, x2-8x + 15 = 2, ∧ 0, the equation has two unequal roots, when y = - 2, x2-8x + 15 = - 2, ∧ 0, the equation has no real root, and ∧ there are two qualified points n, so we choose C

What is the symmetry of the graph of the function f (x) = LG (2 divided by 1-x and then - 1)
Why the origin symmetry

F (x) = LG (2 divided by 1-x and then - 1)
=lg(2/(1-x)-1)
=lg(1+x)/(1-x)
f(-x)=lg(1-x)/(1+x)=-lg (1+x)/(1-x)=-f(x)
-f(x)=f(-x)
Odd function
So it's symmetric about the origin

Given the function f (x) = LG (1-x / 1 + x), the image of function g (x) is axisymmetric with the image of function y = - (1 / x + 2), Let f (x) = f (x) + G (x)
(1) Finding the analytic expression and definition field of function f (x)
(2) Whether there are two different points a and B on the function f (x) image, so that the line AB is just perpendicular to the Y axis, and the coordinates are obtained

F(x)=lg(1-x/1+x)-(1/-x+2),
The second is to prove that f (x) = 0 has more than two roots

Known sequence {an}, {BN} satisfy: A1 = 1 / 4, an + BN = 1, BN + 1 = BN / 1-an ^ 2. Find {BN} general term formula