Question

设二次函数f(x)=x^2+x,当x属于[n,n+1],(n属于正整数)时,f(x)的所有整数值的个数为g(n)

Answer 1

f(n) = n^2 + n

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

之间的整数的个数g(n) = f(n+1)-f(n)+1 = 2n+3

an = (2n^3+3n^2)/(2n+3) = n^2(an+3)/(2n+3) = n^2

所以 sn = 1^2 + 2^2 + 3^2 + …… n^2 = n(n+1)(2n+1)/6