Question

设二次函数f(x)=x^2+x…X属于[n,n+1]n为正整数；求他的整数值g(n)的个数

Answer 1

f(n)=n^2+n

f(n+1)=n^2+2n+1+n+1=n^2+3n+2

显然f(x)在[n,n+1]连续，且为单调递增。故它的整数值g(n)的个数为：

n^2+3n+2-(n^2+n)+1

=2n+3

Answer 2

解：易知f在（0，正无穷）单调增

f(n+1)=(n+1)^2+n+1=n^2+3n+2

f(n)=n^2+n.

x从n变到n+1时，对应的f从n^2+n变化到n^2+3n+2

而这两数之间的整数个数为n^2+3n+2-（n^2+n）+1=2n+1.

即g(n)=2n+1