Sum of sum of binomial coefficients $\sum_{j=1}^{n}{\sum_{k=0}^{j}{{n}\choose{k}}}$ I know there is no simple way to solve the sum:
$$\sum_{k=0}^{j}{{n}\choose{k}}$$
But what about the equation:
$$\sum_{j=1}^{n}{\sum_{k=0}^{j}{{n}\choose{k}}}$$
Are there any simplifications or good approximations for this equation?

Basically it’s just a matter of reversing the order of summation, much as you might reverse the order of integration of a double integral, though the lower limit of $1$ on the outer summation requires a little adjustment.
$$\begin{align*}
\sum_{j=1}^n\sum_{k=0}^j\binom{n}k&=\sum_{j=0}^n\sum_{k=0}^j\binom{n}k-\binom{n}0\\
&=\sum_{k=0}^n\sum_{j=k}^n\binom{n}k-1\\
&=\sum_{k=0}^n(n-k+1)\binom{n}k-1\\
&=(n+1)\sum_{k=0}^n\binom{n}k-\sum_{k=0}^nk\binom{n}k-1\\
&=(n+1)2^n-\sum_{k=0}^nn\binom{n-1}{k-1}-1\\
&=(n+1)2^n-1-n\sum_{k=0}^{n-1}\binom{n-1}k\\
&=(n+1)2^n-1-n2^{n-1}\\
&=(n+2)2^{n-1}-1
\end{align*}$$
