Showing $3^n=\sum_{k=0}^n{(-1)^k\binom{n}{k}4^{n-k}}$. This is Exercise 1.9.3 of F. M. Goodman's "Algebra: Abstract and Concrete".

Show $$3^n=\sum_{k=0}^n{(-1)^k\binom{n}{k}4^{n-k}}.$$

My Attempt:
I tried induction on $n$ as follows.
Base: Try when $n=1$. Then $LHS=3^1=3$ and
$$\begin{align}
RHS&=\sum_{k=0}^1{(-1)^k\binom{1}{k}4^{1-k}} \\
&=(-1)^0\binom{1}{0}4^1+(-1)^1\binom{1}{1}4^0 \\
&=4-1 \\
&=3
\end{align}$$ so the result holds for $n=1$.
Assume the result for $n=r\in\mathbb{N}$. Then $$3^r=\sum_{k=0}^r{(-1)^k\binom{r}{k}4^{r-k}}.$$
Consider when $n=r+1$: I have $3^{r+1}=3\cdot 3^r=3\sum_{k=0}^r{(-1)^k\binom{r}{k}4^{r-k}}$ and have considered
$$\sum_{k=0}^{r+1}{(-1)^k\binom{r+1}{k}4^{r+1-k}}$$ by writing $\binom{r+1}{k}=\binom{r}{k}+\binom{r}{k-1}$ but to no avail.

Start with 
$$ \sum_{k=0}^{r+1} (-1)^k\binom{r+1}{k}4^{r+1-k} = 4^{r+1} + \sum_{k=1}^{r+1} (-1)^k\Bigg( \binom{r}{k} + \binom{r}{k-1}\Bigg)4^{r+1-k}$$
$$ = 4^{r+1} +  \sum_{k=1}^{r+1} (-1)^k\binom{r}{k}4^{r+1-k} +  \sum_{k=1}^{r+1} (-1)^k\binom{r}{k-1}4^{r+1-k}.$$
In the first sum the term $k=r+1$ is zero and you can add in the $k=0$ term back. In the second sum, reindex your $k$ to go from $0$ to $r$. So you'll have
$$  \sum_{k=0}^{r+1} (-1)^k\binom{r+1}{k}4^{r+1-k} =  \sum_{k=0}^{r} (-1)^k\binom{r}{k}4^{r+1-k} +  \sum_{k=0}^{r} (-1)^{k+1}\binom{r}{k}4^{r-k}. $$
Using your induction hypothesis,
$$ = 4*3^{r} - 3^{r} = 3^{r+1}.$$
However a much quicker proof would be to use the binomial theorem:
$$ (x+y)^r = \sum_{k=0}^{r} \binom{r}{k} x^k y^{r-k} $$
with $x=-1$ and $y=4$.
