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?

$$\begin{align}
\sum_{j=\color{red}0}^{n}{\sum_{k=0}^{j}{{n}\choose{k}}}
&=\sum_{k=0}^n\sum_{j=k}^n\binom nk\\
&=\sum_{k=0}^n (n-k+1)\binom n{n-k}\\
&=\sum_{k=0}^n (j+1)\binom nj
&&\text{putting }j=n-k\\
&=\sum_{k=0}^n j\binom nj+\sum_{k=0}^n \binom nj\\
&=n 2^{n-1}+2^n\\
\sum_{j=\color{red}1}^n\sum_{k=0}^n\binom nk&=\sum_{j=\color{red}0}^n\sum_{k=0}^n\binom nk-1\\
&=n 2^{n-1}+2^n-1\\
&=(n+2)2^{n-1}-1\quad\blacksquare
\end{align}$$
