TITLE: Summation by parts of $\sum_{k=0}^{n}k^{2}2^{k}$
QUESTION [12 upvotes]: I want to evaluate this sum
$$\sum_{k=0}^{n}k^{2}2^{k}$$ by summation by parts (two times) and I need to know, if my approach was right.
I know the formula for summation by parts is $$\sum u\Delta v=uv-\sum\left(Ev\right)\Delta u$$ where $E\left(v\left(x\right)\right)=v\left(x+1\right)$ and $(v(x))=v(x+1)-vx)$.
Further if $f$ is a Antiderivative of $g$ I know the formula
$$\sum_{k=a}^{b}g\left(k\right)=f\left(b+1\right)-f\left(a\right)$$
First I choose $u=x^{2}$. This means $\Delta u=\Delta x^{2}=2x+1$. Choose $v=2^{x}$. That means $\Delta v=2^{x}$. Then I get
$$\begin{eqnarray*}
\sum_{k=0}^{n}k^{2}2^{k} &=& \sum_{0}^{n+1}x^{2}2^{x} \\
&=& \left.x^{2}*2^{x}\right|_{0}^{n+1}-\sum_{0}^{n+1}2^{x+1}(2x+1) \\
&=& (n+1)^2 2^{n+1}-\sum_{0}^{n+1}2x2^{x+1}+2^{x+1} \\
&=& (n+1)^2 2^{n+1}-4\sum_{0}^{n+1}x2^{x} - 2\sum_{0}^{n+1}2^{x}
\end{eqnarray*}$$
Now choose $u(x)=x$. This means $\Delta u=\Delta x=1$. Chose $v=2^{x}$. This means $\Delta v=2^{x}$ and $E(v(x))=2^{x+1}$. I get then
$$\begin{eqnarray*}
\sum_{x=0}^{n+1}x2^{x+1}
&=& \left[x2^{x}-2^{x+1}\right]_{0}^{n+1} \\
&=& ((n+1)-1)2^{(n+1)+1}+2 \\
&=& n2^{n+2}+2
\end{eqnarray*}$$
Putting those two calculations together resolves in
$$\begin{eqnarray*}
\sum_{k=0}^{n}k^{2}2^{k}
&=&(n+1)^2 2^{n+1}-2\sum_{0}^{n+1}x2^{x+1} - \sum_{0}^{n+1}2^{x+1} \\
&=& (n+1)^2 2^{n+1}-2(n2^{n+2}+2) - (n+2)2^{x} \\
&=& 2^{n+1}(n^2+2n+1) -4(n2^{n+2}+2) - 2(n+2)2^{x} \\
&=& 2^{n+1}(n^2+2n+1-8n-(n+2))-8 \\
&=& 2^{n+1}(n^2-7n-1)-8 \\
\end{eqnarray*}$$
But Mathematica says
$$2^{n+1} ((n-2) n+3)-6 = 2^{n+1} (n^2-2n+3)-6$$
Where is my fault?
Update
I placed the right calculations as an answer below.

REPLY [7 votes]: I calculated it again by pen and paper after letting a little time passing. And I got it! I made some basic mistakes, which I am ashamed of now, that I see them. But you learn by any *kind* of mistakes, don't ya? ;-)
Here the correct calculation:
First I calculated
\begin{equation*}
\sum_{k=0}^{n} k 2^{k}
\end{equation*}
by summation by parts. I chose $x=u$ and $\Delta 2^x=v$. This gave me $\Delta x=1$ and $v=2^x$. I see the descrete integration of $2^x$ behaves like the analytical integration of $e^x$. (I hope the vocabular I used here is correct, English is not my native language. :-))
$$
\begin{eqnarray*}
\sum_{k=0}^{n} k 2^{k} &=& \sum_{0}^{n+1}x2^{x} \\
&=& \left[x2^{x}\right]_{0}^{n+1}-\sum_{0}^{n+1}2^{x+1}\cdot 1 \\
&=& \left(n+1\right)2^{n+1}-2\sum_{0}^{n+1}2^{x} \\
&=& \left(n+1\right)2^{n+1}-2\left(2^{n+1}-1\right) \\
&=& n2^{n+1}+2^{n+1}-2\cdot2^{n+1}+2 \\
&=& n2^{n+1}-2^{n+1}+2 \\
&=& 2^{n+1}\left(n-1\right)+2
\end{eqnarray*}
$$
Now I can evaluate
\begin{equation*}
\sum_{k=0}^{n} k^2 2^{k}
\end{equation*}
For that I chose $x^2=u$ and $\Delta 2^x=v$. This gave me $\Delta x=2x+1$ and $v=2^x$. Now summation by parts and some basic calculations yields
$$
\begin{eqnarray*}
\sum_{k=0}^{n}k^{2}2^{k} &=& \sum_{0}^{n+1}x^{2}2^{x} \\

&=& \left[x^{2}2^{x}\right]_{0}^{n+1}-\sum_{0}^{n+1}2^{x+1}\left(2x+1\right) \\
&=& \left(n+1\right)^{2}2^{n+1}-\left(\sum_{0}^{n+1}2x2^{x+1}+2^{x+1}\right) \\
&=& \left(n^{2}+2n+1\right)2^{n+1}-\left(4\sum_{0}^{n+1}x2^{x}+2\sum_{0}^{n+1}2^{x}\right)\\
&=& \left(n^{2}+2n+1\right)2^{n+1}-\left(4\left(\left(n-1\right)2^{n+1}+2\right)+2\left[2^{x}\right]_{0}^{n+1}\right) \\
&=& \left(n^{2}+2n+1\right)2^{n+1}-\left(4\left(\left(n-1\right)2^{n+1}+2\right)+2\left(2^{n+1}-1\right)\right) \\
&=& \left(n^{2}+2n+1\right)2^{n+1}-\left(4\left(\left(n2^{n+1}-2^{n+1}\right)+2\right)+2\cdot2^{n+1}-2\right) \\
&=& \left(n^{2}+2n+1\right)2^{n+1}-\left(4\left(n2^{n+1}-2^{n+1}\right)+8+2\cdot2^{n+1}-2\right) \\
&=& \left(n^{2}+2n+1\right)2^{n+1}-\left(4n2^{n+1}-42^{n+1}+8+2\cdot2^{n+1}-2\right) \\ 
&=& \left(n^{2}+2n+1\right)2^{n+1}-4n2^{n+1}+42^{n+1}-8-2\cdot2^{n+1}+2 \\
&=& 2^{n+1}\left(\left(n^{2}+2n+1\right)-4n+4-2\right)-8+2 \\
&=& 2^{n+1}\left(n^{2}-2n+3\right)-6 \\
&=& 2^{n+1}\left((n-2)n+3\right)-6

\end{eqnarray*}
$$
which is exacly what WolframAlpha proclaims. Thanks to anyone who suggested corrections. :-)