A Floor Function Dilemma!

$?=\lim_{ n \to \infty }\frac{1}{n^2}\sum_{k=1}^{n-1}{k\left\lfloor x+\frac{n-k-1}{n}\right\rfloor}\;\;\;\;\;\;\;\;\;\;\;\;0≤\frac{n-k-1}{n}<1\;\;\;\;\;\;\;\;\;\;\;\;\;\{x\}=x-\lfloor x\rfloor$ $Let\;\;\;1= \{x\}+\frac{n-p-1}{n}\;\;\;⇒\;\;\;p=n\{x\}-1\;\;\;and\;\;\;\left\lfloor x+\frac{n-k-1}{n} \right\rfloor=\begin{cases} \lfloor x\rfloor+1\;\;\;if\;\;\;k≤\lfloor p\rfloor≤p\\ \lfloor x\rfloor\;\;\;if\;\;\;k≥\lfloor p\rfloor+1>p\end{cases}$ $?=\lim_{n\to\infty}\frac{1}{n^2}\left((\lfloor x\rfloor+1)\sum_{k=1}^{\lfloor p\rfloor}{k }+\lfloor x\rfloor\sum_{k=\lfloor p\rfloor+1 }^{n-1}{ k}\right) \;\;\;Using\;the\;arithmetic\;seriers\;formula\;\;\;⇒$ $?=\lim_{n\to\infty}\frac{1}{n^2}\left((\lfloor x\rfloor+1)\frac{(\lfloor p\rfloor +1)(\lfloor p\rfloor-1+1 )}{2}+\lfloor x\rfloor\frac{(n-1+\lfloor p\rfloor +1)(n-1-\lfloor p\rfloor -1+1)}{2}\right)$ $If\;\;\;n\to\infty\;\;\;⇒\;\;\;\lfloor p\rfloor\to p\to n\{x\} \to\infty$ $?=\lim_{n\to\infty}\frac{(\lfloor x\rfloor+1)p^2+\lfloor x\rfloor(n^2-p^2)}{2n^2}=\lim_{n\to\infty}\frac{\lfloor x\rfloor n^2+p^2 }{2n^2}=\frac{\lfloor x\rfloor +\{x\}^2}{2}=\frac{44+(\sqrt{2015}-44)^2}{2}\approx\boxed{22.395}$

@Satyajit Mohanty Sir , can you please post a solution, because I got it correct by mistake , I got exactly 22 but the answer was 22.395 . I would like to know the solution. Please sir.