I am trying to find a short form of the long sum $$ \sum_{k=1}^{n-1} (-4)^k \binom{n}{n-k} 3^{n-k} $$,
which I can map somewhat awkwardly to the binomial formula $$ \sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k} = (x+y)^n $$ by changing the bounds, getting $$\sum_{k=0}^{n} \binom{n}{k} (-4)^k * 3^{n-k} = (-4+3)^n = (-1)^n$$
How can I also accurately account for the fact that the original summation starts at $k=1$ and ends at $n-1$, and not $k=0$ and $n$?