Sum of 2 raised to the power of every element in a line of Pascal’s triangle

Question

I know that the elements of a line in Pascal’s triangle add up to $2^n$ . What about:

$$\sum_{k=0}^n 2^{\binom{n}{k}}$$ For example, line $n = 2$ adds up to $8$. $n = 3$ adds up to $20$. Is there any formula?

Answer 1

If you make a new triangle out of pascal triangle by replacing $\binom {n}{k} $ with $2^{ \binom {n}{k} }$ then and the additive property

$$\binom {n+1}{k}= \binom {n}{k} + \binom {n}{k-1} $$ translates to the multiplicative property $$ 2^{ \binom {n+1}{k}} = 2^{\binom {n}{k}}\times 2^{\binom {n}{k-1}}$$

The product of elements on the $n_{th} $ row would be $ 2^{2^n} $

I did not find a simple formula for the sum of rows of the new triangle.