Question

已知f(x)对任意的实数x,y都有f(x+y)=f(x)+f(y)+2y(x+y)=1,且f(1)=1.若x属于正整数，求f(x)的表达式。

Answer 1

f(x+y)=f(x)+f(y)+2y(x+y)+1
令y=1
f(x+1)=f(x)+f(1)+2(x+1)+1=f(x)+2x+4
即
f(x+1)=f(x)+2x+4
同理有
f(x)=f(x-1)+2(x-1)+4
f(x-1)=f(x-2)+2(x-2)+4
......
f(2)=f(1)+2*1+4
f(1)=1
将上面x个式子相加得
f(x)
=1+2*(1+2+...+(x-1))+4*(x-1)
=1＋(1+x-1)x+4x-4
=x^2+4x-3