Why is $5^{n+1}+2\cdot 3^n+1$ a multiple of $8$ for every natural number $n$? I have to show by induction that this function is a multiple of 8. I have tried everything but I can only show that is multiple of 4, some hints? The function is
$$5^{n+1}+2\cdot 3^n+1 \hspace{1cm}\forall n\ge 0$$, because it is a multiple of 8, you can say that$$5^{n+1}+2\cdot 3^n+1=8\cdot m \hspace{1cm}\forall m\in\mathbb{N}$$.

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$\ds{\large n\ \mathsf{even}:}$
\begin{align}
5^{n + 1} + 2 \times 3^{n} + 1 & =
5\pars{5^{n} - 1} + 2\pars{3^{n} - 1} + 8 =
5\pars{4\overbrace{\sum_{k = 0}^{n - 1}5^{k}}^{\ds{even}}}\ +\
4\ \overbrace{\sum_{k = 0}^{n - 1}3^{k}}^{\ds{even}} + 8
\end{align}

$\ds{\large n\ \mathsf{odd}:}$
\begin{align}
5^{n + 1} + 2 \times 3^{n} + 1 & =
\pars{5^{n + 1} - 1} + 6\pars{3^{n - 1} - 1} + 8
\\[5mm] & =
4\overbrace{\sum_{k = 0}^{n}5^{k}}^{\ds{odd:\ 2p + 1}}\ +\
12\ \overbrace{\sum_{k = 0}^{n - 2}3^{k}}^{\ds{odd:\ 2q + 1}} + 8
\qquad\qquad\pars{~p\ \mbox{and}\ q\ \mbox{are}\ integers~}
\\[5mm] & = 8p + 24q + 3 \times 8
\end{align}
