How to solve the summation of series $a^{i}(x+i)$ where $i$ is from $1$ to $N$ I have the following series and I am unable to figure out which series it belongs to and how to solve it
$a(x+1)+a^{2}(x+2)+…+a^{N}(x+N)$
Above series is a generalization of my actual series
$\dfrac{1}{2}(x+1)+\dfrac{1}{4}(x+2) +\cdots +\dfrac{1}{2^{N}}(x+N)$

Read all sums as sums from $0$ to $n$
Using the identity $\sum_k a^k=\frac{1-a^{N+1}}{1-a}$ we can calculate $\sum_k ka^k$
$\sum_k (k+1)a^k=1+a+a^2+...+a^N+a+a^2+....+a^N+a^2+a^3+...+a^N+...+a^N=\frac{1-a^{N+1}}{1-a}+a\frac{1-a^{N}}{1-a}+a^2\frac{1-a^{N-1}}{1-a}+....a^N\frac{1-a}{1-a}=\frac{1+a+a^2+...+a^N - Na^{N+1}}{1-a}=\frac{\frac{1-a^{N+1}}{1-a}-Na^{N+1}}{1-a}$
So $\sum_k ka^k=\sum_k (k+1)a^k-\sum_k a^k=\frac{\frac{1-a^{N+1}}{1-a}-Na^{N+1}-(1-a^{N+1})}{1-a}=\frac{1-Na^{N+1}+(N-1)a^{N+2}}{(1-a)^2}-\frac{1}{1-a}$
