Let a > 0 be a constant. Find a simplified, asymptotically tight bound for the recurrence T(n) = aT(n-2) + C
The first four lines of your argument were correct; all you missed was the proper generalization. In general, we see that
$$
T(n) = c+c\alpha+c\alpha^2+\dotsb+c\alpha^{j-1}+\alpha^j T(n-2j)
$$
Now we want to drive $T(n-2j)$ down to a value we know, namely $T(1)=c$. To do this we'll need $n-2j=1$ and so we set $j=(n-1)/2$, giving us
$$\begin{align}
T(n) &amp;= c+c\alpha+c\alpha^2+\dotsb+c\alpha^{j-1}+\alpha^j T(n-2j)\\
     &amp;= c+c\alpha+c\alpha^2+\dotsb+c\alpha^{(n-1)/2-1}+\alpha^{(n-1)/2} c\\
     &amp;= c(1+\alpha+\alpha^2+\dotsb+\alpha^{(n-1)/2})
\end{align}$$
and this is just a geometric series, so I'll leave the last steps to you.