Question

已知函数f(x)=x^3+bx^2+cx+d,f(x)的三个零点x1,x2,x3分别为公差为3的等差数列｛an｝的前三项

Answer 1

f'(x) / 3 + 3 = x^2 + (2bx + c)/3 + 3 = (x+b/3)^2 - b^2/9 + c/3 + 3 的解唯一，所以b^2 /9 = c/3 + 3

b^2 = 3c + 27

f(x) = (x-x1)(x-x2)(x-x3)

解得c = 3, d = 10, b = -6

x1 = -1, x2= 2, x3 = 5, xn = -1 + (n-1)*3 = 3n - 4

an = 3n-4

n = 20时，Tn=550