Let $S(N,a,k)= \sum_{n=0}^N a^n n^k$. Multiplying by $x^k/k!$ and summing on $k$ gives the exponential generating function
\[\sum_{k=0}^\infty S(N,a,k) \frac{x^k}{k!}= \frac{(ae^x)^{N+1}-1}{ae^x-1}. \]$$\sum_{k=0}^\infty S(N,a,k) \frac{x^k}{k!}=
\frac{(ae^x)^{N+1}-1}{ae^x-1}.$$
From this formula, it is easy to calculate $S(N,a,k)$ for small values of $k$ using Maple, Mathematica, or Sage, etc.
For $a=1$, we have the well known expression for $S(N,a,k)$ as a polynomial in $N$ in terms of Bernoulli numbers. Now suppose that $a\ne 1$. Then $S(N,a,k) = T(N,a,k) - U(a,k),$ where
\[\sum_{k=0}^\infty T(N,a,k) \frac{x^k}{k!}= \frac{(ae^x)^{N+1}}{ae^x-1}\]$$\sum_{k=0}^\infty T(N,a,k) \frac{x^k}{k!}=
\frac{(ae^x)^{N+1}}{ae^x-1}$$
and
\[\sum_{k=0}^\infty U(a,k) \frac{x^k}{k!}= \frac{1}{ae^x-1}.\]$$\sum_{k=0}^\infty U(a,k) \frac{x^k}{k!}=
\frac{1}{ae^x-1}.$$
From the first formula it is not hard to show that
\[T(N,a,k)=\frac{a^{N+1}}{(a-1)^{k+1}} Q_k(N,a) \]$$T(N,a,k)=\frac{a^{N+1}}{(a-1)^{k+1}} Q_k(N,a)$$ where $Q_k(N,a)$ is a polynomial in $N$ and $a$.
Using the well-known formula
${1}/(1-ae^x)= \sum_{k=0}^\infty A_k(a)/(1-a)^{k+1}$, where $A_k(a)$ is an Eulerian polynomial, we see that
$U(a,k)=(-1)^{k} A_k(a)/(a-1)^{k+1}$. The coefficients of powers of $N$ in $Q_k(N,a)$ can also be expressed explicitly in terms of Eulerian polynomials in $a$.
