Find an egf for $\sum^{n}_{k=0} \binom{n}{k}$

Question

The egf would be $\sum_{n = 0} [\sum^{n}_{k=0} \binom{n}{k}]\frac{x^{n}}{n!}$ = $\sum_{n = 0} \frac{n!}{k!(n-k)!}\frac{x^{n}}{n!}$ = $\sum_{n = 0} \frac{x^{n}}{k!(n-k)!}$

From here I'm a little stuck, can someone direct me to some formulas that seem to be eluding me?

Answer 1

We know the exponential generating function of $e^x$ is \begin{align*} e^x=\sum_{n=0}^\infty\frac{x^n}{n!} \end{align*}

Since $e^{2x}=e^xe^x$, we obtain \begin{align*} e^{2x}=e^xe^x&=\left(\sum_{k=0}^{\infty}\frac{x^k}{k!}\right)\left(\sum_{l=0}^\infty\frac{x^l}{l!}\right)\\ &=\sum_{n=0}^{\infty}\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{x^n}{k!l!}\\ &=\sum_{n=0}^{\infty}\left(\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{n!}{k!l!}\right)\frac{x^n}{n!}\\ &=\sum_{n=0}^{\infty}\left(\sum_{k=0}^n\binom{n}{k}\right)\frac{x^n}{n!}\\ \end{align*}