What's the issue?? Handle the case where a = b separately and note that 10^9+9 is prime number so write (a/b) as a*b^(-1) modulo N, N = 10^9+9....Where are you facing problem??

You can apply the same technique that is used for binary exponentiation.

Even case: $$$a^0 + a^1 + \ldots + a^{2k - 1} = (a^0 + a^k) \cdot (a^0 + a^1 + \ldots + a^{k - 1})$$$

Odd case: $$$a^0 + a^1 + \ldots + a^{2k - 1} + a^{2k} = (a^0 + a^k) \cdot (a^0 + a^1 + \ldots + a^{k - 1}) \cdot a + 1$$$, where the last $$$1$$$ is just $$$a^0$$$.

In both cases, we reduce finding the sum until $$$2k$$$ or $$$2k + 1$$$ to finding the sum until $$$k$$$.