If f(n)=f(n−1)+nf(n) = f(n-1) + nf(n)=f(n−1)+n, what is f(n)f(n)f(n) in terms of nnn (assuming f(0)=0f(0)=0f(0)=0)?
f(n)=n(n+1)/2f(n) = n(n+1)/2f(n)=n(n+1)/2
f(n)=n2f(n) = n^2f(n)=n2
f(n)=n(n−1)/2f(n) = n(n-1)/2f(n)=n(n−1)/2
f(n)=2nf(n) = 2nf(n)=2n