Why is $5^{n+1}+2\cdot 3^n+1$ a multiple of $8$ for every natural number $n$?

Question

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$$5^{n+1}+2\cdot 3^n+1=8\cdot m \hspace{1cm}\forall m\in\mathbb{N}$$.

Answer 1

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.

Answer 2

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.

Answer 3

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 $4$:

$5^{k+2}-1 = 4.5^{k+1} + 5^{k+1} -1$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\large n\ \mathsf{even}:}$ \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}

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$\ds{\large n\ \mathsf{odd}:}$ \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}