Even It Out 2

When there are an even number of operators, the last added term was even. If the sum is even, then one of four things can happen to get to the next added even term: $+$ odd $+$ even which makes odd, $+$ odd $\times$ even which makes even, $\times$ odd $+$ even which makes even, and $\times$ odd $\times$ even which makes even. If the sum is odd, then one of four things can happen to get to the next added even term: $+$ odd $+$ even which makes even, $+$ odd $\times$ even which makes odd, $\times$ odd $+$ even which makes odd, and $\times$ odd $\times$ even which makes odd.

Therefore, $3$ even sums will be obtained from a previous even sum, and $1$ even sum will be obtained from a previous odd sum, so that if $e_n$ is the number of even sums and $o_n$ is the number of odd sums for $n$ number of operators, $e_n = 3e_{n - 2} + o_{n - 2}$ . Since the evens and odds add up to a power of $2$ , $e_{n - 2} + o_{n - 2} = 2^{n - 2}$ . Substituting this in gives $e_n = 3e_{n - 2} + (2^{n - 2} - e_{n - 2})$ or $e_n = 2e_{n - 2} + 2^{n - 2}$ , which can inductively be shown to be the explicit equation $e_n = 2^{\frac{n}{2} - 1}(2^{\frac{n}{2}} + 1)$ for $e_0 = 1$ .

For consecutive numbers $2$ through $30$ , there are $n = 30 - 2 = 28$ operators. Therefore, the number of even sums is $e_{28} = 2^{\frac{28}{2} - 1}(2^{\frac{28}{2}} + 1) = \boxed{134225920}$ .