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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 $$\sum_{k=0}^\infty S(N,a,k) \frac{x^k}{k!}= \frac{(ae^x)^{N+1}-1}{ae^x-1}.$$ 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 $$\sum_{k=0}^\infty T(N,a,k) \frac{x^k}{k!}= \frac{(ae^x)^{N+1}}{ae^x-1}$$ and $$\sum_{k=0}^\infty U(a,k) \frac{x^k}{k!}= \frac{1}{ae^x-1}.$$ From the first formula it is not hard to show that $$T(N,a,k)=\frac{a^{N+1}}{(a-1)^{k+1}} Q_k(N,a)$$ 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 $a$.