Find the general formula for the partial sum? $\sum 2^{(n-1)} \cdot n$ from $n=1$ to $k$

Question

I'm having a bit of trouble finding the general formula for the following partial sum

$$ \sum_{n=1}^k 2^{n-1} \cdot n $$

Answer 1

Isn't this just $(\frac{d}{dx}\{1 +x+x^2+\cdots + x^k\})|_{x=2}$ ? The series sums to $(1-x^{k+1})/(1-x)$, so its derivative is $\frac{-(k+1)x^k(1-x)+(1-x^{k+1})}{(1-x)^2}$, which, evaluated at $x=2$, simplifies to $(k+1)2^k+1-2^{k+1}=(k+1-2)2^k +1=\boxed{(k-1)2^k+1}$.

Answer 2

Setting $$ S_k=\sum_{n=1}^k2^{n-1}n, $$ we have \begin{eqnarray} S_k&=&\sum_{n=1}^{k}2^{n-1}n=\sum_{n=0}^{k-1}2^n(n+1)=\sum_{n=0}^{k-1}2^nn+\sum_{n=0}^{k-1}2^n=2\sum_{n=0}^{k-1}2^{n-1}n+\frac{1-2^k}{1-2}\\ &=&2^k-1+2\sum_{n=1}^k2^{n-1}n-2\cdot2^{k-1}k=(1-k)2^k-1+2S_k, \end{eqnarray} it follows that $$ S_k=(k-1)2^k+1 $$

Answer 3

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.