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

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?

• Welcome to MSE! I've added MathJax to make your post clearer. You might have a look here for future posts. – Bill O'Haran Jun 27 '18 at 12:30
• Thanks. I am typing in a phone and didn’t know how to do that – user1000 Jun 27 '18 at 12:31
• First few terms are $2,4,8,20,80,2116,\ldots$ OEIS knows not of such a sequence – gt6989b Jun 27 '18 at 12:41
• @gt6989b: Looks like user1000 needs to submit a new sequence to the OEIS! I find it remarkable that it's not there. – Adrian Keister Jun 27 '18 at 12:45
• Fourth term should be $100$ not $80$. Sequence A001315 in OEIS: oeis.org/A001315 – gandalf61 Jun 27 '18 at 13:38

## 1 Answer

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.