Deriving a formula for the coefficients of a power series. Here is the image containing the question

Consider the two  truncated  power  series $u(t)=u_0+u_1x+u_2x^2+u_3x^3+u_4x^4+\cdots+u_nx^n$ and $d(t)=d+d_1x+d_2x^2+d_3x^3+d_4x^4+\cdots+dx^n$.  Show  how to derive  the     formula
  \begin{align*}
q_i=\frac{\sum_{k=0}^{i-1}q_kd_{i-k}}{d_0}
\end{align*}
  where  the $q_i$ are  the coefficients  of the truncated  power series for $q(t)=\frac{u(t)}{d(t)}$.

I have tried rearranging and rewriting and many other things. Help would be greatly appreciated!
 A: 
Hint: The statement of this problem shows some serious flaws. 
  
  
*
  
*$u(t)=u_0+u_1x+u_2x^2+u_3x^3+u_4x^4+\cdots+u_nx^n$:  The left-hand side indicates a function in $t$ while the right-hand side indicates a function in $x$.
  
*$d(t)=d+d_1x+d_2x^2+d_3x^3+d_4x^4+\cdots+dx^n$: The same flaw as above. Additionally, the first and last coefficient are $d$ but should presumably be $d\color{blue}{_0}$  and  $d\color{blue}{_n}$ which  are different  values in general.
  
*$q_i=\frac{\sum_{k=0}^{i-1}q_kd_{i-k}}{d_0}$:  The range of $i$  is missing here. The formula for $q_i$ seems to be incorrect.

Usually we would make the  approach
\begin{align*}
q(x)=q_0+q_1x+q_2x^2\cdots+q_nx^n=\frac{u(x)}{d(x)}
\end{align*}
from which we derive
\begin{align*}
&u_0+u_1x+u_2x^2+\cdots+u_nx^n\\
&\qquad=(q_0+q_1x+q_2x^2+\cdots+q_nx^n)(d_0+d_1x+d_2x^2+\cdots d_nx^n)
\end{align*}
We now compare terms with equal powers, analyse
\begin{align*}
u_0&=q_0d_0\\
u_1&=q_0d_1+q_1d_0\\
u_2&=q_0d_2+q_1d_1+q_2d_0\\
&\vdots\\
u_n&=q_0d_n+q_1d_{n-1}+\cdots+q_nd_0
\end{align*}
from which another formula follows.
