Question

（文）已知向量 a =(x2+1,-x),b =(1,2 n2+1 ) （n为正整数）,函数

Answer 1

a =(x^2+1,-x),b =(1,2n^2+1 ) （n为正整数）,

f(x)=ab=x^2+1-(2n^2+1)x,

(1)依题意an=n^2+1/2.

(2)猜bn=(a+an)(a-an)=[(n+1)^2+1/2+n^2+1/2][(n+1)^2-n^2]

=(2n^2+2n+2)(2n+1),

求bn的前n项和,似乎超出中学数学范围.

(3)kij=(ai^2-aj^2)/(i-j)=(i^2-j^2)/(i-j)=i+j,

当i=2008,j=2010时,直线AiAj的斜率kij=2008+2010=4018.