Sum of the sequence What is the sum of the following sequence
$$\begin{align*}
(2^1 - 1) &+ \Big((2^1 - 1) + (2^2 - 1)\Big)\\
&+ \Big((2^1 - 1) + (2^2 - 1) + (2^3 - 1) \Big)+\ldots\\
&+\Big( (2^1 - 1)+(2^2 - 1)+(2^3 - 1)+\ldots+(2^n - 1)\Big)
\end{align*}$$
I tried to solve this. I reduced the equation into the following equation
$$n(2^1) + (n-1)\cdot2^2 + (n-2)\cdot2^3 +\ldots$$
but im not able to solve it further. Can any one help me solve this equation out. and btw its not a Home work problem. This equation is derived from some puzzle.
Thanks in advance
 A: We have 
$$\begin{align}
\sum_{i=1}^n\sum_{j=1}^i(2^j-1)
&=\sum_{i=1}^n\sum_{j=1}^i2^j-\sum_{i=1}^n\sum_{j=1}^i1\\
&=\sum_{i=1}^n(2^{i+1}-2)-\sum_{i=1}^ni\\
&=2\sum_{i=1}^n2^{i}-\sum_{i=1}^n2-\sum_{i=1}^ni\\
&=2(2^{n+1}-2)-2n-\frac12n(n+1)\\
&=2^{n+2}-4-\frac52n-\frac12n^2\\
\end{align}
$$
A: Let's note that $$(2^1 - 1) + (2^2 - 1) + \cdots + (2^k - 1)  = 2^{k+1} - 2-k$$ where we have used the geometric series.  Thus, the desired sum is actually $$\sum_{k=1}^n{2^{k+1}-2-k}$$.  As this is a finite sum, we can evaluate each of the terms separately.  We get the sum is $$2\left(\frac{2^{n+1}-1}{2-1}-1\right) - 2n- \frac{n(n+1)}{2} = 2^{n+2}-4 - 2n-\frac{n(n+1)}{2} $$
A: Others have given the correct answer; here’s how you could have simplified your incorrect expression.
$$\begin{align*}
n(2^1) + (n-1)\cdot2^2 + (n-2)\cdot2^3 +\ldots&=\sum_{k=1}^n(n-k+1)2^k\\
&=(n+1)\sum_{k=1}^n2^k-\sum_{k=1}^nk2^k\\
&=(n+1)\left(2^{n+1}-2\right)-\sum_{k=1}^n\sum_{i=1}^k2^k\\
&=(n+1)\left(2^{n+1}-2\right)-\sum_{i=1}^n\sum_{k=i}^n2^k\\
&=(n+1)\left(2^{n+1}-2\right)-\sum_{i=1}^n\left(2^{n+1}-2^i\right)\\
&=(n+1)\left(2^{n+1}-2\right)-n2^{n+1}+\sum_{i=1}^n2^i\\
&=(n+1)\left(2^{n+1}-2\right)-n2^{n+1}+2^{n+1}-2\\
&=2\cdot2^{n+1}-2n-4\\
&=2^{n+2}-2n-4
\end{align*}$$
