<s>[INST] Find the general formula for the partial sum? $\sum 2^{(n-1)} \cdot n$ from $n=1$ to $k$ I'm having a bit of trouble finding the general formula for the following partial sum
$$ \sum_{n=1}^k 2^{n-1} \cdot n $$
 [/INST] First note that $2^{n-1}+2^{n-1}=2^n$.  By using this formula for different values of $k$  we can derive a formula for the sum without the $n$ factor:
\begin{eqnarray*}
2^{k} &=& 2^{k-1} + 2^{k-1}\\
&=& 2^{k-1} + 2^{k-2} + 2^{k-2}\\
&=& 2^{k-1} + 2^{k-2}+2^{k-3}+2^{k-3} \\
&=& \cdots\\
&=& (2^{k-1}+2^{k-2}+\cdots+1) + 1\\
&=& 1+\sum_{n=1}^k 2^{n-1}
\end{eqnarray*}
and therefore $\sum_{n=1}^k 2^{n-1} = 2^{n}-1$.
Now we can introduce the factor of $n$.
\begin{eqnarray*}
\sum_{n=1}^k n2^{n-1} &=& 1 + 2*2^1 + 3*2^2 + 4*2^3 + \cdots +n 2^{n-1}\\
&=& 1 \\
&& +2^1 + 2^1 \\
&& +2^2+2^2+2^2\\
&& + \vdots \\
&& +2^{k-1}+2^{k-1}+2^{k-1}+\cdots + 2^{k-1}
\end{eqnarray*}
Now sum each of the columns of the last line.  The first column is 
$1+2^1+2^2+\cdots+2^{k-1}=2^k-1$, the second column is $2^1+2^2+\cdots+2^{k-1}=2^k-2$
and the $i$'th column is $2^i+2^{i+1}+\cdots+2^{k-1}=2^k-2^i$.  Therefore the whole sum can also be written
\begin{eqnarray*}
\sum_{n=1}^k n2^{n-1} &=& (2^k-1)+(2^k-2^1)+(2^k-2^2)+\cdots + (2^k-2^{k-1})\\
&=& k 2^k - \sum_{n=1}^{k}2^{n-1}\\
&=& k2^k-(2^k-1)\\
&=& 1+(k-1)2^k\\
\end{eqnarray*}
This is the most elementary derivation I can think of.  There are faster and simpler methods as well.  Hopefully this was helpful to you.
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