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
$\dfrac{1}{2}(x+1)+\dfrac{1}{4}(x+2) +\cdots +\dfrac{1}{2^{N}}(x+N)$
 A: Hint:
Note that
$$
\begin{align}
(1-a)\sum_{k=1}^na^k
&=\sum_{k=1}^na^k-\sum_{k=1}^na^{k+1}\\
&=\sum_{k=1}^na^k-\sum_{k=2}^{n+1}a^k\\
&=a-a^{n+1}\tag{1}
\end{align}
$$
Therefore,
$$
\sum_{k=1}^na^k=\frac{a-a^{n+1}}{1-a}\tag{2}
$$
Similarly note that
$$
\begin{align}
(1-a)\sum_{k=1}^nka^k
&=\sum_{k=1}^nka^k-\sum_{k=1}^nka^{k+1}\\
&=\sum_{k=1}^nka^k-\sum_{k=2}^{n+1}(k-1)a^k\\
&=\sum_{k=1}^na^k-na^{n+1}\\
&=\frac{a-a^{n+1}}{1-a}-na^{n+1}\\
&=\frac{a-(n+1)a^{n+1}+na^{n+2}}{1-a}\tag{3}
\end{align}
$$
Therefore,
$$
\sum_{k=1}^nka^k=\frac{a-(n+1)a^{n+1}+na^{n+2}}{(1-a)^2}\tag{4}
$$
Formulas $(2)$ and $(4)$ should help.
A: We write the series as

$$ \sum_{k=1}^{N} (x+k)a^k = x\sum_{k=1}^{N} {a^k}+ \sum_{k=1}^{N} {k}{a^k} \longrightarrow (1). $$

You can use the identity

$$ \sum_{k=1}^{N} t^k = \frac{t^{N+1}-t}{t-1}  \longrightarrow (*)$$

to sum the series in $(1)$.
Added: The first series in $(1)$ is straightforward just substitute $t=a$ in $(*)$ as
$$ x\sum_{k=1}^{N} {a^k} = x\frac{a^{N+1}-a}{a-1} .$$
For the second series you just need to differentiate $(*)$ as
$$  \sum_{k=1}^{N} kt^{k-1} = \frac{d}{dt}\frac{t^{N+1}-t}{t-1}. $$
Multiplying the above equation by $t$ yields 

$$ \sum_{k=1}^{N} kt^{k} = t\frac{d}{dt}\frac{t^{N+1}-t}{t-1}. $$

It is your job to do the differentiation and then substitute $t=a$. 
A: 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 $\sum_k ka^k=\sum_k (k+1)a^k-\sum_k a^k=\frac{\frac{1-a^{N+1}}{1-a}-Na^{N+1}-(1-a^{N+1})}{1-a}=\frac{1-Na^{N+1}+(N-1)a^{N+2}}{(1-a)^2}-\frac{1}{1-a}$
