Evaluating $\sum_{k=0}^n \binom{n}{k} 2^{k^2}$

Solution 1:

I agree with J.M.'s comment above that this is not going to resolve into something simple. Asymptotically, though, the last term will dominate, and maybe that will be useful. We have $$2^{n^2} \leq \sum_{k=0}^n \binom{n}{k} 2^{k^2} \leq 2^{n^2}\left(1 + \frac{2n^2}{4^n}\right).$$ The relative remainder term $R(n) = \frac{2n^2}{4^n}$ goes to $0$ fairly quickly as $n \to \infty$, and so $\sum_{k=0}^n \binom{n}{k} 2^{k^2} \approx 2^{n^2}$.

Derivation: The terms in the sum are strictly increasing. Since $n \geq k$, for $k \geq 1$ we have $$\frac{\binom{n}{k} 2^{k^2}}{\binom{n}{k-1}2^{(k-1)^2}} = 2^{2k-1}\left(\frac{n}{k}-1+\frac{1}{k}\right) \geq 2^{2k-1}\left(1-1+\frac{1}{k}\right) = \frac{2^{2k}}{2k}> 1.$$

Therefore, $$2^{n^2} \leq \sum_{k=0}^n \binom{n}{k} 2^{k^2} = 2^{n^2} + \sum_{k=0}^{n-1} \binom{n}{k} 2^{k^2} \leq 2^{n^2} + \sum_{k=0}^{n-1} \binom{n}{n-1} 2^{(n-1)^2} = 2^{n^2} + n^2 2^{n^2-2n+1}$$ $$= 2^{n^2}\left(1 + \frac{2n^2}{4^n}\right).$$