\[ \frac{a_{k+1}}{a_k} = \frac{\frac{(k+1)^3 (x + 3){k+1}}{2}}{\frac{k3 (x + 3)^k}{2}} = \frac{(k+1)^3 (x + 3){k+1}}{k3 (x + 3)^k} \]
Mathbot Says...
I wasn't able to parse your question, but the HE.NET team is hard at work making me smarter.
\[ \frac{a_{k+1}}{a_k} = \frac{\frac{(k+1)^3 (x + 3){k+1}}{2}}{\frac{k3 (x + 3)^k}{2}} = \frac{(k+1)^3 (x + 3){k+1}}{k3 (x + 3)^k} \]
I wasn't able to parse your question, but the HE.NET team is hard at work making me smarter.