Ink on Sequences #2

From the first condition, we see that:

$\begin{aligned} &\sum_{k=1}^{n-1} (a_{2k+1}-a_{2k-1})=\sum_{k=1}^{n-1} (2k-1) \\ &a_{2n-1}-a_{1}=n(n-1)-(n-1) \\ &\therefore~a_{2n-1}=n^2+pn+q \end{aligned}$

And from the second condition, we see that:

$\begin{aligned} &\prod_{k=1}^{n-1} \frac{a_{2k+2}}{a_{2k}}=\prod_{k=1}^{n-1} k \\ &\frac{a_{2n}}{a_2}=(n-1)! \\ &\therefore~a_{2n}=a_2\cdot(n-1)! \end{aligned}$

Also, $a_6=2a_4=2a_2$ .

Punch those into the expression we're trying to find a value of.

$\begin{aligned} &\lim_{n\rightarrow\infty}\frac{12a_{2n}a_{2n+9}}{a_6\cdot(n+1)!} \\ &=\lim_{n\rightarrow\infty}\frac{12a_2\cdot(n-1)!\cdot\{(n+5)^2+p(n+5)+q\}}{2a_2\cdot(n+1)!} \\ &=\lim_{n\rightarrow\infty}\frac{6(n^2+rn+s)}{n(n+1)} \\ &=\boxed{6} \end{aligned}$