how to find formula for the summation series I need a formula for the summation of the series   $$
n+2(n−1)+2^2(n−2)+2^3(n−3)\cdots $$

Is there any way to simplify it further? I am guessing it will be of form
$$
\sum_{j=0}^{n} (n-j)a^j
$$
For the above summation series i have:  
$
n + 2(n-1) + 4(n-2) + 8(n-3)
$

$
2^k(n-k) + \sum_{j=0}^{k-1} (n-j)a^j
$

n-k is zero so

$
2^k + n(2^ -1) + \sum_{j=0}^{n-1} ja^j
$

I am just not able to get it into geometric series form.
 A: You can combinatorially prove that $$\sum_{j=0}^n (n-j)2^j = 2^{n+1} - n - 2$$
by counting subsets $A$ of $\{1,\dots,n+1\}$ with $|A|\ge 2$.  For the right-hand side, exclude the $\binom{n+1}{0}+\binom{n+1}{1}=1+(n+1) = n+2$ subsets of size at most one.  For the left-hand side, condition on the second smallest element $n+1-j$.  Then choose the smallest element from among $\{1,\dots,n-j\}$ in $n-j$ ways, and choose an arbitrary subset of $\{n-j+2,\dots,n+1\}$ in $2^j$ ways.
A: Looking at the terms, cannot help but think of geometric series and differentiation.
$$
\begin{align}
f(x)&=2^{n-1}\sum_{j=0}^{n}{}x^{n-j}\\
&=2^{n-1}\cdot\frac{x^{n+1}-1}{x-1}
\\
\\
\sum_{j=0}^{n}{(n-j)2^{j}}&=\frac{d}{dx}f\left(\frac{1}{2}\right)\\
&=2^{n+1}-n-2
\end{align}
$$
A: Your sum is:
$\begin{align*}
  S_n
    &= \sum_{0 \le k \le n} 2^k (n - k) \\
    &= n \sum_{0 \le k \le n} 2^k - \sum_{0 \le k \le n} k \cdot 2^k
\end{align*}$
First one is just a geometric sum:
$\begin{align*}
  \sum_{0 \le k \le n} 2^k
    &= \frac{2^{n + 1} - 1}{2 - 1} \\
    &= 2^{n + 1} - 1
\end{align*}$
Second one can be computed by a trick. Start with:
$\begin{align*}
   \sum_{0 \le k \le n} z^k
     &= \frac{1 - z^{n + 1}}{1 - z} \\
   z \frac{\mathrm{d}}{\mathrm{d} z} \sum_{0 \le k \le n} z^k
     &= z \frac{\mathrm{d}}{\mathrm{d} z}  \frac{1 - z^{n + 1}}{1 - z} \\
   \sum_{0 \le k \le n} k z^k
     &=
\end{align*}$
Evaluate the derivative on the right hand side at $z = 2$ to get $(2 n - 2) 2^n + 2$. Pull both together:
$\begin{align*}
  \sum_{0 \le k \le n} 2^k (n - k)
    &= n \cdot (2^{n + 1} - 1) - (2 n - 2) \cdot 2^n - 2 \\
    &= 2^{n + 1} - n - 2
\end{align*}$
