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.
 A: 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:
$$\sum_{b=1}^{t-1} \binom{t}{b} \cdot 2^{t-b}=\sum_{b=0}^{t} \binom{t}{b} \cdot 1^{b}\cdot 2^{t-b}-\binom{t}{0} \cdot 1^0 \cdot 2^{t-0}-\binom{t}{t} \cdot 1^{t} \cdot 2^{t-t} \\ =(1+2)^t-\frac{t!}{0!t!}\cdot 1^0 \cdot 2^t-\frac{t!}{t!0!}\cdot 1^t \cdot 2^0=3^t-2^t-1 $$
A: Expand the right hand side: write $3^t$ as $(2+1)^t$ and use the binomial expansion.
A: This may be viewed as two ways of counting the number of pairs $(A,B)$ with $A \subseteq B$ and $A \neq \varnothing$ and $A \neq \{1,\dots,t\}$.    On the left hand side we condition on size.  First, pick your set $A$, the number of ways is $\binom{t}{b}$.  Next, pick its superset $B$ which we can do in $2^{t-b}$ ways.  On the right hand side, we first consider all pairs $(A,B), A \subseteq B$, including letting $A = \varnothing$ or $\{1,\dots,t\}$.  The number of such pairs is $3^t$ which follows from the fact we are picking whether each element of $\{1,2,\dots,t\}$ belongs to $A$, $B \setminus A$ or $B^c$.  The $2^t+1$ we subtract are the $2^t$ pairs coming from $A = \varnothing$ and the one pair from $A =\{1,\dots,t\}$.
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\begin{align}
\color{#66f}{\large\sum_{b = 1}^{t - 1}{t \choose b}2^{t - b}}&=
2^{t}\bracks{-1 + \sum_{b = 0}^{t}{t \choose b}\pars{\half}^{b} - 2^{-t}}
=
2^{t}\bracks{-1 + \pars{1 + \half}^{t}- 2^{-t}}
\\[3mm]&=
2^{t}\bracks{-1 + \pars{3 \over 2}^{t}- 2^{-t}}
=\color{#66f}{\large-2^{t} + 3^{t} - 1}
\end{align}
