More than 2 resistors in parallel
Yes, it is, but the easiest and most compact representation is the 

$${1\over R_t} = \sum_{i=1}^n {1\over R_i} = {1\over R_1} + \dots + {1\over R_n} $$

you already mentioned. To see the pattern you can do the math yourself for 3 resistors:

$${1\over R_t} = {1\over R_1} + {1\over R_2} + {1\over R_3} $$

Do the first addition using the usual method:

$${1\over R_t} = {R_1 + R_2\over R_1R_2} + {1\over R_3}$$

Then do the second addition, using the same method:

$${1\over R_t} = {R_1R_2 + R_2R_3 + R_1R_3\over R_1R_2R_3}$$

Solved for \$R_t\$:

$${R_t} = {R_1R_2R_3\over R_1R_2 + R_2R_3 + R_1R_3}$$

As you can see, the denominator is the sum of the products of "each with each" resistor value, which is more complicated to write than just

$${1\over R_t} = {1\over R_1} + {1\over R_2} + {1\over R_3} $$