f(n) = f(n-1) + (n+1)
f(1)=2, so,
f(2) = f(1) + 3 = 2 + 3
f(3) = f(2) + 4 = 2+3+4
f(4) = f(3) + 5 = 2+3+4+5
...
f(n) = ... = 2+3+4+5+...+n+(n+1)
=(1+2+3+4+5+...+n)+n
=[n(n+1)/2]+n
=[n(n+1)+2n]/2
=n(n+3)/2
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