帮忙解一道数列题

用符号 ^2 表示平方

f（x+3)+f(x-3)=2x^2 -10x+26
f(x+3)-f(x-3）=12x-20
两式相加
2 f(x+3) = 2x^2 + 2x + 6
f(x+3)
= x^2 + x + 3
= (x+3 -3)^2 + (x+3)
所以
f(x) = (x-3)^2 + x = x^2 -5x + 9

f(x-2) = (x-2)^2 - 5(x-2) + 9 = x^2 - 9x + 23
f(5) + 4 = 5^2 - 5*5 + 9 = 9
f(x+2) = (x+2)^2 - 5(x+2) + 9 = x^2 - x + 3

三者成等插数列，所以
f(x-2) + f(x+2) = 2 [ f(5) + 4]
2x^2 - 10x + 26 = 18
x^2 - 5x + 4 = 0
(x-1)(x-4) = 0
x = 1 或 4

该等差数列递增，所以
f(x+2) - f(x-2) > 0
(x^2 - x + 3) - (x^2 - 9x + 23) > 0
8x - 20 > 0
x > 20/8

所以 x = 1 舍去
而 x = 4

f(x-2) = f(2) = 2^2 - 2*5 + 9 = 3
f(x+2) = f(6) = 6^2 - 6*5 + 9 = 15

所以等差数列前3项为 3 9 15，公差为6
An = a1 + (n-1)*d = 3 + (n-1)*6 = 6n -3

Sn = (A1 + An)*n/2 = (3 + 6n -3)*n/2 = 3n^2