Why is $\sum\limits_{b=1}^{t-1} {t \choose b} 2^{t-b} = (3^t - 2^t - 1)$ Why is $$\sum\limits_{b=1}^{t-1} {t \choose b} 2^{t-b} = (3^t - 2^t - 1)$$
Thanks.

Using the binomial theorem:
$$(x+y)^n=\sum_{k=0}^{n} \binom{n}{k} \cdot x^{k} \cdot y^{n-k}$$
we have the following:
$$\sum_{b=1}^{t-1} \binom{t}{b} \cdot 2^{t-b}=\sum_{b=0}^{t} \binom{t}{b} \cdot 1^{b}\cdot 2^{t-b}-\binom{t}{0} \cdot 1^0 \cdot 2^{t-0}-\binom{t}{t} \cdot 1^{t} \cdot 2^{t-t} \\ =(1+2)^t-\frac{t!}{0!t!}\cdot 1^0 \cdot 2^t-\frac{t!}{t!0!}\cdot 1^t \cdot 2^0=3^t-2^t-1 $$
