Summation of n(2^n) I was doing a question in which i had to find the summation of the expression $n(2^n)$ from n=1 to n=9. 
I used wolfram alpha to calculate thid sum, but i was wondering if there is an easier way to calculate it? 
 A: This is an "arithmetico–geometric" progression (a product of an AP
with a GP) and there are standard methods to sum these.
Let
$$S_m=\sum_{k=1}^n k2^k=2+2\times 2^2+\cdots+m\times 2^m.$$
Then
\begin{align}
S_m&=2S_m-S_m=2^2+2\times 2^3+\cdots+m\times 2^{m+1}
-(2+2\times 2^2+\cdots+m\times 2^m)\\
&=-2-2^2-\cdots-2^m+m2^{m+1}=2-2^{m+1}+m2^{m+1}=(m-1)2^{m+1}+2
\end{align}
using the formula for the sum of a GP.
A: Let's assume n is finite. We will now derive an explicit formula for calculating the desired sum for any $x$, you can plug in $2$ into final formula.  The sum: $S_1=\sum_{k=0}^{n} kx^{k}$ looks a lot like: $S_2=\sum_{k=0}^{n} x^{k}$. $S_2$ is of course $\mathbb{geometric}$ series: $S_2 = \frac{1-x^{n+1}}{1-x}$. Since n is finite, we can safely take the derivative of $S_2$: $S_3 = \frac{d}{dx}(\sum_{k=0}^{n} x^{k})=\sum_{k=0}^{n} kx^{k-1}=\frac{d}{dx}(\frac{1-x^{n+1}}{1-x})=...=\frac{-nx^{n}+nx^{n+1}-x^{n}+1}{(1-x)^2}$.
We wanted a sum of $kx^k, x = 2$ and we derived explicit formula for sum of $kx^{k-1}$. We now just need multiply $S_3$ by $x$: $xS_3 = x\sum_{k=0}^{n} kx^{k-1} = \sum_{k=0}^{n} kx^{k} = S_1 = x\frac{-nx^{n}+nx^{n+1}-x^{n}+1}{(1-x)^2} = \frac{-nx^{n+1}+nx^{n+2}-x^{n+1}+x}{(1-x)^2}$.
For $x=2$ we get: $\sum_{k=0}^{n} k2^{k} = -n2^{n+1}+n2^{n+2}-2^{n+1}+2 = 2^{n+1}(n-1)+2$
A: We will start by introducing the geometric progression summation formula:
$$\sum_{i=a}^b c^i = \frac{c^{b-a+1}-1}{c-1}\cdot c^{a}$$
Finding the sum of series $\sum_{i=1}^{n}i\cdot b^{i}$ is still an unresolved problem, but we can very often transform an unresolved problem to an already solved problem. In this case, the geometric progression summation formula will help us.
Steps:
$$
\sum_{i=1}^{n}i\cdot b^{i} = 
\sum_{i=0}^{n-1}\sum_{j=n-i}^{n}b^{j}=
\sum_{i=0}^{n-1}\frac{b^{\left(n-\left(n-i\right)+1\right)}-1}{b-1}\cdot b^{\left(n-i\right)}=
\sum_{i=0}^{n-1}\frac{b^{n+1}-b^{n-i}}{b-1}=
\frac{\sum_{i=0}^{n-1}b^{n+1}-\sum_{i=0}^{n-1}b^{n-i}}{b-1}=
\frac{nb^{n+1}-\sum_{i=0}^{n-1}b^{i+1}}{b-1}=
\frac{nb^{n+1}-\frac{b^{n}-1}{b-1}\cdot b}{b-1}=
\frac{nb^{n+1}}{b-1}-\frac{b^{n+1}-b}{\left(b-1\right)^{2}}
$$
Now, just replace $b$ with $2$
$$
\frac{n\cdot2^{n+1}}{2-1}-\frac{2^{n+1}-2}{\left(2-1\right)^{2}}=
\left(n-1\right)2^{n+1}+2
$$
