Question

f(x+y)=f(x)+f(y)+2y(x+y)+1 ,f(1)=1,当X属于正整数时,求f(x)的解析式?

Answer 1

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

y=1

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

=f(x)+2x+4

f(x+1)=f(x)+2x+4

f(n)=An

A(n+1)=A(n)+2n+4

A(n)=A(n-1)+2(n-1)+4

……

A(2)=A(1)+2+4

A(1)=1

叠加

A(n+1)=1+4n+2(1+2+……n)

＝1＋4n＋n(1+n)

=-3+4(n+1)+n(n+1)

f(n)=A(n)=n(n-1)+4n-3