Question:

Step 1: Evaluate the Numerator Let the numerator be $N$. $$N = \sum_{k=1}^{n-1}(k-1)(nk-k^2)$$ $$N = \sum_{k=1}^{n-1}(nk^2 - k^3 - nk + k^2)$$ $$N = \sum_{k=1}^{n-1}((n+1)k^2 - k^3 - nk)$$ Using Faulhaber's formulas for sums up to $m=n-1$: $$\sum_{k=1}^{n-1} k = \frac{(n-1)n}{2}$$ $$\sum_{k=1}^{n-1} k^2 = \frac{(n-1)n(2n-1)}{6}$$ $$\sum_{k=1}^{n-1} k^3 = \left(\frac{(n-1)n}{2}\right)^2$$ Substitute these into the expression for $N$: $$N = (n+1)\frac{(n-1)n(2n-1)}{6} - \left(\frac{(n-1)n}{2}\right)^2 - n\frac{(n-1)n}{2}$$ $$N = \frac{n(n^2-1)(2n-1)}{6} - \frac{n^2(n-1)^2}{4} - \frac{n^2(n-1)}{2}$$ Expanding and collecting terms for large $n$: $$N = \frac{n(2n^3-n^2-2n+1)}{6} - \frac{n^2(n^2-2n+1)}{4} - \frac{n^3-n^2}{2}$$ $$N = \frac{2n^4-n^3-2n^2+n}{6} - \frac{n^4-2n^3+n^2}{4} - \frac{n^3-n^2}{2}$$ The leading term is obtained from the $n^4$ terms: $$N = \left(\frac{2}{6} - \frac{1}{4}\right)n^4 + O(n^3)$$ $$N = \left(\frac{1}{3} - \frac{1}{4}\right)n^4 + O(n^3)$$ $$N = \frac{1}{12}n^4 + O(n^3)$$ Step 2: Evaluate the Denominator Let the denominator be $D$. $$D = 2\sum_{r=1}^n r^3-\sum_{s=0}^n(s^2+(n-s)^2)$$ First part: $$2\sum_{r=1}^n r^3 = 2\left(\frac{n(n+1)}{2}\right)^2 = 2\frac{n^2(n+1)^2}{4} = \frac{n^2(n^2+2n+1)}{2} = \frac{n^4+2n^3+n^2}{2}$$ Second part: $$\sum_{s=0}^n(s^2+(n-s)^2) = \sum_{s=0}^n(s^2+n^2-2ns+s^2)$$ $$= \sum_{s=0}^n(2s^2-2ns+n^2)$$ $$= 2\sum_{s=0}^n s^2 - 2n\sum_{s=0}^n s + \sum_{s=0}^n n^2$$ Using Faulhaber's formulas for sums up to $n$: $$\sum_{s=0}^n s^2 = \frac{n(n+1)(2n+1)}{6}$$ $$\sum_{s=0}^n s = \frac{n(n+1)}{2}$$ $$\sum_{s=0}^n n^2 = (n+1)n^2$$ Substitute these into the expression: $$\sum_{s=0}^n(s^2+(n-s)^2) = 2\frac{n(n+1)(2n+1)}{6} - 2n\frac{n(n+1)}{2} + (n+1)n^2$$ $$= \frac{n(n+1)(2n+1)}{3} - n^2(n+1) + n^2(n+1)$$ $$= \frac{n(n+1)(2n+1)}{3}$$ $$= \frac{n(2n^2+3n+1)}{3} = \frac{2n^3+3n^2+n}{3}$$ Now combine the parts for $D$: $$D = \frac{n^4+2n^3+n^2}{2} - \frac{2n^3+3n^2+n}{3}$$ $$D = \frac{3(n^4+2n^3+n^2) - 2(2n^3+3n^2+n)}{6}$$ $$D = \frac{3n^4+6n^3+3n^2 - 4n^3-6n^2-2n}{6}$$ $$D = \frac{3n^4+2n^3-3n^2-2n}{6}$$ The leading term is $\frac{3}{6}n^4 + O(n^3) = \frac{1}{2}n^4 + O(n^3)$. Step 3: Calculate the limit $t$ $$t = \lim_{n\to\infty}\dfrac{N}{D} = \lim_{n\to\infty}\dfrac{\frac{1}{12}n^4 + O(n^3)}{\frac{1}{2}n^4 + O(n^3)}$$ $$t = \dfrac{1/12}{1/2} = \dfrac{1}{12} \cdot 2 = \dfrac{1}{6}$$ Step 4: Calculate $[43t]$ $$[43t] = \left[43 \cdot \frac{1}{6}\right] = \left[\frac{43}{6}\right] = [7.166...] = 7$$