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