So f (2) = f (1)+f (1)+2 *1*1= 4.

3= 1+2

So f (3) = f (1)+f (2)+2 *1* 2 = 9.

f(4)= f( 1)+f(3)+2 * 1 * 3 = 16

So f (n) = n 2

certificate

N= 1, which is obviously correct.

If n=k, k>= 1

That is, f (k) = k 2

Then n=k+ 1.

f(k+ 1)=f(k)+f( 1)+2*k* 1=k^2+ 1+2k=(k+ 1)^2

To sum up, when n is an integer,

f(n)=n^2