You have the correct recurrence relation a_{2n} = - \frac{2n+1}{n+1}a_n. You just haven't tried to solve it.
Recurrences of the form <br />
a_{n+1} = f(n)a_n have the solution <br />
a_n = a_0\prod_{k=0}^{n-1} f(k) where by convention <br />
\prod_{k=0}^{-1} f(k) = 1.
Recurrences of the form <br />
a_{n+2} = f(n)a_n can be turned into the above form by treating even and odd terms separately: First set n = 2m and b_m = a_{2m} to obtain <br />
b_{m+1} = f(2m)b_m and then set n = 2m+1 and c_m = a_{2m+1} to obtain <br />
c_{m+1} = f(2m+1)c_m.