A problem of sum floors let $n$ be a positive integer, prove that $$\sum_{i=0}^{\left\lfloor\frac{n}{3}\right\rfloor}\left\lfloor\frac{n-3i}{2}\right\rfloor=\left\lfloor\frac{n^2+2n+4}{12}\right\rfloor.$$
It looks like we have to consider lots of cases to solve the problem. Are there any simple solutions?
 A: I think this is an easy way to handle it. We induct on $n$. See we use a step argument. Call 
$$S_n=\sum_{i=0}^{\lfloor{n/3}\rfloor}\left\lfloor{\frac{n-3i}{2}}\right\rfloor$$
We have to prove $S_n=\left\lfloor{\dfrac{(n+1)^2+3}{12}}\right\rfloor \tag{1}$
Now $S_{n+3}-S_{n}=\left\lfloor{\dfrac{n+3}{2}}\right\rfloor$ To prove $(1)$ we need to prove:
$$ \left\lfloor{\dfrac{(n+1)^2+3}{12}}\right\rfloor+\left\lfloor{\dfrac{n+3}{2}}\right\rfloor=\left\lfloor{\dfrac{(n+4)^2+3}{12}}\right\rfloor\tag{2}$$
To prove $(2)$ we replace $n$ by $n-1$ and 
$$\left\lfloor{\dfrac{n^2+3}{12}}\right\rfloor+\left\lfloor{\dfrac{n+2}{2}}\right\rfloor-\left\lfloor{\dfrac{(n+3)^2+3}{12}}\right\rfloor=Q_n$$
Now we show 
$$Q_{n+6}=Q_n\tag{3}$$ which is a consequence of trivial algebra. Hence to prove $(2)$ for all $n$ we only need prove it for $n=0,1,2,3,4,5$ due to $(3)$. Which is trivial. Now to prove $(1)$ for all $n$ we need to check $n=0,1,2$ and we are again done. Note I haven't done all the calculations, they are left for the OP, but I have checked they are true indeed. I suppose this answers your questions.
