Let the function f (x) = (x-3) 3+x- 1, {an} is a arithmetic progression with a tolerance other than 0, f (a1)+f (a2)+...+f () a7 =14, and then a.

Analysis: ∫{ an} is a arithmetic progression with a tolerance of not 0, f (a1)+f (a2)+...+f () a7 =14.

∴[(a 1-3)^3+a 1- 1]+[(a2-3)^3+a2- 1]+…+[(a7-3)^3+a7- 1]= 14

∵ The function h (x) = x 3 is a odd function, which is symmetric about the origin center.

∴ h (x-3) = (x-3) 3, which is symmetric about the center of point (3,0).

∫{ an} is an arithmetic series, and the tolerance is not 0.

∴h(a 1)+h(a2)+.....+h(a7)=0

∴ points (A 1, H (A 1)), (A2, H (A2)), ... and (A7, H (A7)), (A6, H (A6)), ...

(a4，h(a4)=(3，0)

∴(a 1-3)^3+[(a2-3)^3+…+[(a7-3)^3=0

∴[(a 1-3)^3+a 1- 1]+[(a2-3)^3+a2- 1]+…+[(a7-3)^3+a7- 1]= 14

a 1- 1+a2- 1+…+a7- 1 = 14

a 1+a2+…a7 = 7+ 14 = 2 1