Limit of Sums

$\lim_{n\rightarrow\infty}{\frac{\sum\limits_{k=1}^n{k^2(n+1-k)}}{\sum\limits_{k=1}^n{k^3}}}\\ =\lim_{n\rightarrow\infty}{\frac{\sum\limits_{k=1}^n{k^2(n+1)-\sum\limits_{k=1}^n{k^3}}}{\sum\limits_{k=1}^n{k^3}}}\\ =\lim_{n\rightarrow\infty}{\frac{(n+1)\sum\limits_{k=1}^n{k^2}}{\sum\limits_{k=1}^n{k^3}}-1}$ Now we can use the formulas for the sum of the first n squares and cubes: $\begin{aligned}\sum\limits_{k=1}^n{k^2}&=\frac{n(n+1)(2n+1)}{6}&&&\sum\limits_{k=1}^n{k^3}&=\frac{n^2(n+1)^2}{4}\end{aligned}$ $\begin{aligned}\lim_{n\rightarrow\infty}{\frac{(n+1)\sum\limits_{k=1}^n{k^2}}{\sum\limits_{k=1}^n{k^3}}-1}&=\lim_{n\rightarrow\infty}{\frac{(n+1)\frac{n(n+1)(2n+1)}{6}}{\frac{n^2(n+1)^2}{4}}-1}\\ &=\lim_{n\rightarrow\infty}{\frac{4n(2n+1)}{6n^2}-1}\\ &=\lim_{n\rightarrow\infty}{\frac{4n^2+2n}{3n^2}-1}\\ &=\lim_{n\rightarrow\infty}{\frac{4+\frac{2}{n}}{3}-1}\\ &=\frac{4}{3}-1\\ &=\frac{1}{3}\end{aligned}$ So the final answer is $2(1)+3(3)=\boxed{11}$

$\begin{aligned} L & = \lim_{n \to \infty} \frac {\sum_{k=1}^n k^2(n+1-k)}{\sum_{k=1}^n k^3} \\ & = \lim_{n \to \infty} \frac {(n+1)\sum_{k=1}^n k^2-\sum_{k=1}^n k^3}{\sum_{k=1}^n k^3} \\ & = \lim_{n \to \infty} \frac {(n+1)\times \frac {n(n+1)(2n+1)}6}{\frac {n^2(n+1)^2}4} - 1 \\ & = \lim_{n \to \infty} \frac {4(2n+1)}{6n} - 1 \\ & = \lim_{n \to \infty} {\color{#3D99F6}\frac {4n+2}{3n}} - 1 & \small \color{#3D99F6} \text{Divide up and down by }n \\ & = \lim_{n \to \infty} \frac {4+\frac 2n}3 - 1 \\ & = \frac 43 - 1 = \frac 13 \end{aligned}$

Therefore, $2a+3b = 2(1)+3(3) = \boxed{11}$ .