Quad Function Sum

Let's first calculate $S_N = \sum_{i = 1}^{N^4-1} \frac 1{\lfloor \sqrt[4]{i}\rfloor} = \sum_{k = 1}^{N-1} \frac{(k+1)^4 - k^4} k \\ = \sum_{k=1}^{N-1} \left(4k^2 + 6k + 4 + \frac 1 k\right) = \tfrac23 (N-1)N(2N-1) + 3(N-1)N + 4(N-1) + \sum_{k=1}^{N-1} \frac 1 k.$ Since $6^4-1 = 1295 < 2016 < 7^4-1 = 2400$ , we evaluate this for $N = 6$ , and add $2016 - 1295 = 721$ times the term $1/6$ : $S = S_6 + 721\times\tfrac16 = \tfrac23 (6-1)6(2\cdot 6-1) + 3(6-1)6 + 4(6-1) + \sum_{k=1}^5 \frac 1 k + 120\tfrac 16 \\ = \tfrac23\cdot 5\cdot 6\cdot 11 + 3\cdot 5\cdot 6 + 4\cdot 5 + (1 + \tfrac12 + \tfrac13 + \tfrac 14 + \tfrac15) + 120\tfrac16 \\ = 220 + 90 + 20 + 2\tfrac{17}{60} + 120\tfrac 16 = 452\tfrac{27}{60} = 452\tfrac{9}{20} = \frac{9049}{20}.$ The answer is $9049 + 20 = \boxed{9069}$ .

It's a beautiful sum. Because between 1 and 2016 there are only six forth powers, I see no need to use a more general approach. Simple table will do the job. My solution follows: