Solving a recurrence equation that yields polynomials I am trying to solve the following recurrence equation:
$$
T(n) = kT(n - 1) + nd
$$
I have expanded the first 4 values ($n = 1$ was given):
$$\begin{align}
T(1) & = 1 \\
T(2) & = kT(2-1) + 2d = k + 2d \\
T(3) & = kT(3-1) + 3d = k(k + 2d) + 3d = k^2 + 2kd + 3d  \\
T(4) & = kT(4-1) + 4d = k(k^2 + 2kd + 3d) + 4d = k^3 + 2k^2d + 3dk + 4d\\
\end{align}$$
I was able to convert the above into the following summation, in hopes of finding a closed-form solution:
$$
T(n) = k^{n - 1} + d \sum_{i=1}^{n-1} (i + 1) (k^{n-i-1})
$$
But after this point I am stuck. Was creating the summation the correct thing to do, or is there something better I could try?

Using
\begin{align}
\sum_{j=0}^{m} t^{j} = \frac{1-t^{m+1}}{1-t}
\end{align}
then
\begin{align}
\sum_{j=0}^{m} j \, t^{j} = \frac{t - (m+1) t^{m+1} + m t^{m+2}}{(1-t)^{2}}
\end{align}
for which
\begin{align}
\sum_{j=0}^{m} (j+1) \, t^{j} &= \frac{1}{(1-t)^{2}} \left[ (1-t)(1-t^{m+1}) + t - (m+1) t^{m+1} + m t^{m+2} \right] \\
&= \frac{1}{(1-t)^{2}} \left[ 1 - (m+2) t^{m+1} + (m+1) t^{m+2} \right].
\end{align}
Now, 
\begin{align}
\sum_{j=1}^{m} (j+1) t^{j} = \frac{1}{(1-t)^{2}} \left[ 1 - (m+2) t^{m+1} + (m+1) t^{m+2} \right] - 1.
\end{align}
This leads to
\begin{align}
\sum_{j=1}^{m} (j+1) t^{m-j} = \frac{1}{(1-t)^{2}} \left[ t^{m} - (m+2) t + (m+1)  \right] - t^{m}.
\end{align}

Now for the difference equation
\begin{align}
T(n) = k T(n-1) + n d
\end{align}
where $T(1) = 1$ leads to the solution
\begin{align}
T(n) = k^{n-1} + d \sum_{j=1}^{n-1} (j+1) k^{n-j-1} 
\end{align}
which can be seen in the form
\begin{align}
T(n) &= (1-d) k^{n-1} + \frac{d}{(1-k)^{2}} \left[ k^{n-1} - (n+1) k + n \right]
\end{align}
