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$$5^{n+1}+2\cdot 3^n+1=8\cdot m \hspace{1cm}\forall m\in\mathbb{N}$$.
 A: Hint: the difference between consecutive terms is:
$$
\require{cancel}
5^{n+2}+2\cdot 3^{n+1}+ \bcancel{1} - (5^{n+1}+2\cdot 3^n+\bcancel{1})
 = 5^{n+1}(5-1)+2\cdot 3^n(3-1) = 4 \cdot (5^{n+1}+3^n)
$$
The last factor is a sum of two odd numbers.
A: HINT:
If  $a_n=(2b+1)^n$
$$a_{m+2}-a_m=8(2b+1)^m\cdot\dfrac{b(b+1)}2$$ which is multiple of $8$ as $b(b+1)$ is even.
A: By induction:
It is true for case $n=1$, Let it be true for $n=k$ then 
$5^{k+2} +2.3^{k+1} +1 = 5.5^{k+1}+3.2.3^k+1 = 2(5^{k+1} -1) + 3(5^{k+1}+2.3^k+1)$
The second part is a multiple of $8$ and one can easily  show that $(5^{k+1} -1)$ is a multiple of $4$.
Hint for showing  $(5^{k+1} -1)$ is a multiple of $4$:
$5^{k+2}-1 = 4.5^{k+1} + 5^{k+1} -1$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\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}
