Find a close formula using generating functions I want to find the sum:
$$\sum \limits_{k=0}^n {n\choose k}2^{k-n}$$
I did it by using the binomial theorem and I got 
$$\sum \limits_{k=0}^n {n\choose k}2^{k-n}
= 2^{-n} \sum \limits_{k=0}^n {n \choose k}2^k
= \left(\frac{1}{2}\right)^n3^n$$
I would like to know how to do it using generating functions. I have tried some calculations but I am stuck. This is what I have so far.
$$\sum \limits_{k=0}^n {n\choose k}2^{k-n}
=\sum \limits_{n\ge0} \sum \limits_{k\ge0} {n \choose k} 2^{k-n} x^n
=\sum \limits_{k\ge0} 2^k \sum \limits_{n\ge0} {n \choose k}\left(\frac{x}{2}\right)^n$$

The binomial coefficient in $\sum \limits_{k=0}^n \binom{n}{k}2^{k-n}$ indicates that we could use the Cauchy-product of exponential generating functions. We have
\begin{align*}
A(x)=\sum_{k=0}^\infty a_k\frac{x^k}{k!},\ \ B(x)=\sum_{l=0}^\infty b_l\frac{x^l}{l!}\quad\longleftrightarrow\quad
A(x)B(x)=\sum_{n=0}^\infty \left(\sum_{k=0}^n\binom{n}{k}a_kb_{n-k}\right)\frac{x^n}{n!}
\end{align*}

We obtain
  \begin{align*}
\sum_{n=0}^\infty\left(\color{blue}{\sum_{k=0}^n\binom{n}{k}2^{k-n}}\right)\frac{x^n}{n!}
&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{n}{k}1^k\left(\frac{1}{2}\right)^{n-k}\right)\frac{x^n}{n!}\\
&=\left(\sum_{k=0}^\infty\frac{x^k}{k!}\right)\left(\sum_{l=0}^\infty\frac{\left(\frac{x}{2}\right)^l}{l!}\right)\\
&=e^xe^{\frac{x}{2}}\\
&=e^{\frac{3}{2}x}\\
&=\sum_{n=0}^\infty\color{blue}{\left(\frac{3}{2}\right)^n}\frac{x^n}{n!}
\end{align*}
  and the claim follows.

Besides the more common method above we  also have the possibility to use ordinary generating functions. We can apply a transformation known as Euler transform of a series
\begin{align*}
A(x)=\sum_{n=  0}^\infty a_nx^n\quad\longrightarrow\quad
\frac{1}{1-x}A\left(\frac{x}{1-x}\right)=\sum_{n= 0}^\infty \left(\sum_{k=0}^n\binom{n}{k}a_k\right)x^n
\end{align*}
This transformation formula together with a proof can be found e.g.  in Harmonic Number Identities Via Euler's transform by K.N. Boyadzhiev (see Lemma 1).

We obtain
  \begin{align*}
\sum_{n=0}^\infty\left(\color{blue}{\sum_{k=0}^n\binom{n}{k}2^{k-n}}\right)x^n
&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{n}{k}2^k\right)\left(\frac{x}{2}\right)^n\\
&=\frac{1}{1-\frac{x}{2}}\sum_{n=0}^\infty2^n\left(\frac{\frac{x}{2}}{1-\frac{x}{2}}\right)^n\\
&=\frac{1}{1-\frac{x}{2}}\cdot\frac{1}{1-2\left(\frac{\frac{x}{2}}{1-\frac{x}{2}}\right)}\\
&=\frac{1}{1-\frac{3}{2}x}\\
&=\sum_{n=0}^\infty\color{blue}{\left(\frac{3}{2}\right)^n}x^n
\end{align*}
  and the claim follows.

