<s>[INST] Exact closed form expression of $(2^0+...+2^n)+(2^1+...+2^{n+1})+...+(2^n+...+2^{2n})$ Exact closed form of this expression $(2^0+...+2^n)+(2^1+...+2^{n+1})+...+(2^n+...+2^{2n})$
I assume this means there is just one $2^0$ and one $2^{2n}$ and a double of all the terms in between?
 [/INST] Your series is
$1+2+\cdots+2^n$
$\,\,\,\,\,\,\,\,\,2+\cdots+2^n+2^{n+1}$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2^n+2^{n+1}+\cdots+2^{2n}$
And Sum can be written as
$$\begin{align}s&=1+2(2+4+\cdots+2^{n})+2^{n+1}+2^n+2^{n+1}+2^{n+2}+\cdots2^{2n}\\
&=1+2(2+4+\cdots+2^{n})+2^{n+1}+(2^n+2^{n+1}+2^{n+2}+\cdots2^{2n})\\
\end{align}$$
So, answer is

$$1+2\cdot\frac{2^{n+1}-2}{2-1}+2^{n+1}+\frac{2^{2n+1}-2^{n}}{2-1}=2^{2n+1}+3\cdot2^{n+1}-2^n-3$$

</s>