$ \newcommand{\cn}{\colon} \newcommand{\<}{\leqslant} \newcommand{\>}{\geqslant} \newcommand{\ss}{\subset} \newcommand{\k}{\mathrm{k}} \newcommand{\gr}{\mathrm{gr}} \newcommand{\R}{\mathrm{R}} \newcommand{\ov}{\overline} \newcommand{\l}{\lambda} $Let $\k$ be a field, $A$ be associative algebra with unity $1$ over $\k$ and filtration $0=F_{-1}\ss F_0\ss F_1\ss\ldots\ss A=\bigcup_i F_i$, $1\in F_0$, $\gr(A)=\bigoplus_{i\>0} F_i/F_{i-1}$ be associated graded algebra which is finitely generated by homogeneous elements $\ov{a_1},\ldots,\ov{a_n}$ of degrees $d_1,\ldots,d_n$. That is, for all $\ov{f_m}\in F_m/F_{m-1}$ exist $\l_i\in\k,~k_{i,m,j}\in\mathbb N$ such that \begin{align} &F_m/F_{m-1}\ni\ov{f_m}=\sum_{(i,m)} \l_{i,m}\ov{a_{i_{m,1}}^{k_{i,m,1}}\ldots a_{i_{m,n}}^{k_{i,m,n}}}\in\bigoplus_{i\>0} F_i/F_{i-1} \\ \rightsquigarrow~&\forall i,m\cn\ov{a_{i_{m,1}}^{k_{i,m,1}}\ldots a_{i_{m,n}}^{k_{i,m,n}}}\in F_m/F_{m-1} \\ \rightsquigarrow~&\forall i,m\cn (\ov{a_{i_{m,1}}^{k_{i,m,1}}\ldots a_{i_{m,n}}^{k_{i,m,n}}}\ne \ov 0\rightsquigarrow k_{i,m,1}d_{i_{m,1}}+\ldots +k_{i,m,n}d_{i_{m,n}}=m).\tag{1} \end{align}
Def. Rees algebra $\R(A)$ is a subalgebra $\sum_{m\>0}F_m t^m$ of $A[t]$.
I want to prove that $\R(A)$ is finitely generated by $t,a_1t^{d_1},\ldots,a_nt^{d_n}$, where $a_i$ is a preimage of $\ov{a_i}$ (exercise 5-(ii) after chapter 1 from Kassel's Quantum Groups).
Let $\sum f_mt^m\in \R(A)$. Therefore \begin{align} &\sum_m f_mt^m=f_0(a_1t^{d_1})^0+\sum_{m>0}\sum_{(i,m)}\l_{i,m}a_{i_{m,1}}^{k_{i,m,1}}\ldots a_{i_{m,n}}^{k_{i,m,n}}t^m. \end{align} Assume that $\ov{e=a_{i_{m,1}}^{k_{i,m,1}}\ldots a_{i_{m,n}}^{k_{i,m,n}}}=\ov 0\in\gr(A)$. Then $ e\in\cap_{i\>0}F_i=F_0$ and $m=\deg e=0$. As a consequence, from $(1)$ we get $$ \sum_m f_mt^m=f_0(a_1t^{d_1})^0+\sum_{m>0}\sum_{(i,m)}\l_{i,m}(a_{i_{m,1}}t^{d_{i_{m,1}}})^{k_{i,m,1}}\ldots (a_{i_{m,n}}t^{d_{i_{m,n}}})^{k_{i,m,n}}. $$ So $\R(A)$ is generated by $\{a_it^{d_i}\}_{i=1}^n$.
Is it correct?