Finding sum of
\begin{align}
\sum_{0\leq i<j\leq n}\binom{n}{i} \tag1\\
\sum_{0\leq i<j\leq n}\binom{n}{j} \tag2
\end{align}
For $(1)$, we can write it as
\begin{align}
\sum_{0\leq i<j\leq n}\binom{n}{i} 1^j &= 1^1\binom{n}{0}+1^2\left[\binom{n}{0}+\binom{n}{1}\right]+1^3\left[\binom{n}{0}+\binom{n}{1}+\binom{n}{2}\right]+\cdots+\left[\binom{n}{0}+\binom{n}{1}+\cdots +\binom{n}{n-1}\right] \\
&=n\binom{n}{0}+(n-1)\binom{n}{1}+(n-2)\binom{n}{2}+\cdots +\binom{n}{n-1}
\end{align}
How do I solve it after that?
Best Answer
We have $$\sum_{0\leq i<j\leq n}\binom{n}{i}=$$ $$n\binom{n}{0}+(n-1)\binom{n}{1}+(n-2)\binom{n}{2}+\cdots +\binom{n}{n-1}=$$ $$\sum_{i=0}^{n}(n-i)\binom{n}{i}= \sum_{i=0}^{n}(n-i)\binom{n}{n-i}= \sum_{k=0}^{n}k\binom{n}{k}=n2^{n-1},$$ see, for instance, here for the last equality.
Also
$$\sum_{0\leq i<j\leq n}\binom{n}{j}=\sum_{j=0}^{n}j\binom{n}{j}=n2^{n-1}.$$