Find the value of $\underset{0\leq i \lt j\leq n}\sum{\binom{n}{i}}$

Question

My approach:

$$\underset{0\leq i \lt j\leq n}\sum{\binom{n}{i}}=\sum_{j=0}^n\sum_{i=0}^n\binom{n}{i}=\sum_{j=0}^n2^n=2^n(n+1).$$

But the answer is actually $n2^{n-1}$. Am I doing the double sum wrong?

EDIT:

I messed up the indices in the double sum. Following @5xum's answer we have, $$\sum_{j=0}^n\sum_{i=0}^{j-1}{n\choose i}=\sum_{j=0}^n\bigg[\binom{n}{0}+\biggl[\binom{n}{0}+\binom{n}{1}\biggr]+\biggl[\binom{n}{0}+\binom{n}{1}+\binom{n}{2}\biggr]+\cdots$$ $$=\sum_{r=0}^n(n-r)\binom{n}{r}=n2^{n-1}$$ Is this the correct way to do it or is there any other way?

Answer 1

$i$ must be smaller than $j$, and neither of which is the condition missing when you write out the double sum.

So it should be

$$\sum_{0\leq i<j\leq n}{n\choose i} = \sum_{j=0}^n\sum_{i=0}^{j-1}{n\choose i}$$

Answer 2

Your first equality is wrong: you should be summing over $0\leq i<j\leq n$, not $0\leq i,j\leq n$.

$$ \sum_{0\leq i<j\leq n} \binom{n}{i} = \sum_{i=0}^n \sum_{j=i+1}^n \binom{n}{i} $$