I'm trying to find the value of $\sum_{i=0}^{100}\lfloor i^{3/2}\rfloor+\sum_{j=0}^{1000}\lfloor i^{2/3}\rfloor$.
The second sum is obvious, since it's over $j$ instead of $i$. I thought there would be cancellation in the first sum. But it really confuses me how to deal with the square roots inside the floor function.
Any idea is appreciated.
The presence of functions inverse to each other (I mean, $x\mapsto x^{3/2}$ and $x\mapsto x^{2/3}$; look also at the limits) indicates that the "$i$" in the second sum is almost certainly a typo, making one compute the first sum mechanically. With "$j$" in place, it has a somewhat "nice" solution at least. Namely, to compute $$A_n=\sum_{i=0}^{n^2}\lfloor i^{3/2}\rfloor+\sum_{j=0}^{n^3}\lfloor j^{2/3}\rfloor,$$ denote $S_n=\{(i,j) : 1\leqslant i\leqslant n^2,1\leqslant j\leqslant n^3\}$ and observe that the 1st (resp. 2nd) sum is equal to the number of pairs $(i,j)\in S_n$ such that $i^3\geqslant j^2$ (resp. $i^3\leqslant j^2$). Thus, $A_n$ counts each element $(i,j)\in S_n$ once if $i^3\neq j^2$ and twice if $i^3=j^2$; the latter happens if and only if $i=k^2$ and $j=k^3$ for some $k$, $1\leqslant k\leqslant n$. Hence, $$A_n=n^5+n.$$