I'm asked to prove that a sequence of the form $v_n = a \cdot b^{cn+d}, n\in \mathbb{N},\:a,\:b,\:c\:\in \mathbb{R}\backslash \left\{0\right\},\:d\in \mathbb{R} $ is a geometric series of common ratio $b^c$.
I tried to state that $a \cdot b^{cn+d} = a \cdot b^{cn} \cdot b^d$. Since $b^d$ and $a$ are real numbers, their product is a real number (let's call it $c$). So, $a \cdot b^{cn+d} = c \cdot (b^c)^n$.
I'm self-taught, so I have no clue if this is the right way to prove this... It doesn't seem so.
The common ratio is the ratio of successive terms i.e. $v_{n+1} / v_n$
$$ \frac{v_{n+1}}{v_{n}} = \frac{a \, b^{c (n+1) + d}}{ \ a \, b^{c n + d}} = \frac{ b^{c (n+1) + d}}{ b^{c n + d}} = \frac{ b^{c n + c} \, b^d}{ b^{c n} \, b^ d} = \frac{ b^{c n + c} }{ b^{c n} } = \frac{ b^{c n} b^c }{ b^{c n} } = b^c $$