Question

己知f(x)满足f(1)=1,对于任意实数x,y都满足f(x+y)=f(x)+f(y)+2y(x+y)+1,若x属于N,

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+3

即

f(x+1)=f(x)+2x+3

同理有

f(x)=f(x-1)+2(x-1)+3

f(x-1)=f(x-2)+2(x-2)+3

.

f(2)=f(1)+2*1+3

f(1)=1

将上面x个式子相加得

f(x)

=1+2*(1+2+...+(x-1))+3*(x-1)

=x^2+2x-2