I would like to determine the values of $a$ for which $3 \cdot 5^a \cdot 7$ is abundant.
My work so far: $\sigma(3 \cdot 5^a \cdot 7) > 2 \cdot 3 \cdot 5^a \cdot 7 = 42 \cdot 5^a \Leftrightarrow$ $ \sigma(3) \cdot \sigma (5^a) \cdot \sigma (7) > 42 \cdot5^a \Leftrightarrow$
$(4) \cdot \left ( \sum_{k = 0}^a 5^k\right ) \cdot (8) > 42 \cdot 5^a$ ... And since the sum contains $5^a$ in it, I thought about trying to cancel that from both sides, but am stuck.
Can I get a nudge in the right direction? (Also, if there is a theoretic result that I should be using, feel free to mention it!)
Added: Using Will Jagy's hint, I now have $ 8 \cdot (5^{a + 1} - 1) = 40 \cdot 5^a - 8 > 42 \cdot 5^a$ which appears to have no solution.