$f(x_{1},x_{2},...,x_{n}) = a_{0}+a_{1}x_{1}+...+a_{n}x_{n}$ such that $\forall _{i=0,1,2,..,n}$ $a_{i}, x_{i} $ $\epsilon$ {$0,1$}
In simple words $a_{i}, x_{i} $ take boolean values.
Function $f$ take n boolean variables as input and gives a boolean value as output as per the eqaution.
For better understanding lets go from small
$n=1$
| $x_{1}$ |
$f(x_{1}) = a_{0}+a_{1}x_{1}$ |
$f_{1}$
$(a_{0}=0,a_{1}=0 )$ |
$f_{2}$
$(a_{0}=0,a_{1}=1 )$ |
$f_{3}$
$(a_{0}=1,a_{1}=0 )$ |
$f_{4}$
$(a_{0}=1,a_{1}=1 )$ |
| 0 |
$a_{0}$ |
0 |
0 |
1 |
1 |
| 1 |
$a_{0}+a_{1}$ |
0+0 = 0 |
0+1=1 |
1+0=1 |
1+1=0 |
Notice in the truth table that for $x_{1} = 1$, $f(x_{1}) = a_{0}+a_{1}$
we get 4 functions because $a_{0},a_{1}$ takes boolean value.($a_{0},a_{1}$ each can 2 boolean values and hence by product rule total count of number functions $a_{0}$ XOR $a_{1}$ is $2^{2}$).
We get $2^{(1+1)}$ linear boolean functions when $n=1$
$a_{0}+a_{1}$ also covers all possibilities of functions possible from values of $a_{0}$
Extending this to n variables $(x_{1},x_{2},...,x_{n})$ will get $2^{n}$ rows in truth table in which for values of $(x_{1}=1,x_{2}=1,...,x_{n}=1)$ at last row results in $f(x_{1},x_{2},...,x_{n}) = a_{0}+a_{1}+...+a_{n}$.
There will be $n+1$ $a_{i}$ terms in the last row of the truth table and hence number of possible boolean functions is $2^{(n+1)}$
(Notice that question indirectly asked for number of $n+1$-ary boolean functions possible)
