Tìm ${{y}^{(100)}}$ của hàm số $y=\frac{2x+1}{x-1}$

Tìm ${{y}^{(100)}}$ của hàm số $y=\frac{2x+1}{x-1}$
Đặt $u=(2x+1)$ và $v={{(x-1)}^{-1}}$
Ta có ${{y}^{(n)}}=\sum\limits_{k=0}^{n}{C_{n}^{k}{{u}^{(k)}}{{v}^{(n-k)}}}$
Tính được $$\begin{align}
& C_{n}^{0}=1 \\
& C_{n}^{1}=n \\
& C_{n}^{2}=\frac{n!}{(n-2)!.2!}=\frac{1}{2}n(n-1) \\
& C_{n}^{3}=\frac{n!}{(n-3)!.3!}=\frac{1}{6}n.(n-1).(n-2) \\
\end{align}$$
$$\begin{align}
& u=(2x+1)\Rightarrow {{u}^{(0)}}=2x+1 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{u}^{(1)}}=2 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{u}^{(2)}}=0 \\
\end{align}$$
$$\begin{align}
& v={{(x-1)}^{-1}}=>{{v}^{(0)}}={{(x-1)}^{-1}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{v}^{(1)}}=(-1){{(x-1)}^{-2}}={{(-1)}^{1}}{{(x-1)}^{-(1+1)}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{v}^{(2)}}=(-1)(-2).{{(x-1)}^{-3}}=1.2.{{(x-1)}^{-3}}={{(-1)}^{2}}1.2.{{(x-1)}^{-(2+1)}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=>{{v}^{(n)}}={{(-1)}^{n}}n!{{(x-1)}^{-(n+1)}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{v}^{(n-1)}}={{(-1)}^{n-1}}(n-1)!{{(x-1)}^{-n}} \\
\end{align}$$
$$\begin{align}
& \Rightarrow {{y}^{(100)}}=(2x+1){{(-1)}^{100}}100!{{(x-1)}^{-101}}+100.2.{{(-1)}^{99}}99!{{(x-1)}^{-99}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=100!\frac{2x+1}{{{(x-1)}^{101}}}-2.100!\frac{1}{{{(x-1)}^{99}}} \\
\end{align}$$