Proving that $12^n + 2(5^{n-1})$ is a multiple of 7 for $n\geq 1$ by induction

Question

Prove by induction that $12^n + 2(5^{n-1})$ is a multiple of $7$.

Here's where I am right now:

Assume $n= k $ is correct: $$12^k+2(5^{k-1}) = 7k.$$

Let $n= k+1 $: $$12^{k+1} + 2(5^k)$$ $$12^k(12) + 2(5^k)$$

Any ideas on how I can proceed from here?

Answer 1

Basis Step:

$P(1) = 12^1 + 2\cdot5^{1-1} = 12 + 2 = 14 = 7\cdot 2$

Thus $P(1)$ is true.

Inductive Step:

Assume $P(k)$ holds for an arbitrary positive integer $k$. Under this assumption, let us prove that $P(k+1)$ is true, namely that

$$12^{k+1} + 2\cdot5^{k} $$ is also a multiple of 7.

\begin{align} 12^{k+1}+2\cdot 5^k &=12\cdot 12^k+2\cdot 5^k\\ &=12\cdot(7q-2\cdot 5^{k-1})+2\cdot 5^k\\ &=7\cdot12q-24\cdot5^{k-1}+10\cdot5^{k-1}\\ &=7\cdot12q+(-24+10)\cdot5^{k-1}\\ &=7\cdot12q+(-14)\cdot5^{k-1}\\ &=7\cdot12q+(7\cdot(-2))\cdot5^{k-1}\\ \end{align}

$7\cdot12q$ is a multiple of 7. Furthermore, $7\cdot(-2))\cdot5^{k-1}$ is a multiple of 7, thus their sum is a multiple of 7, illustrated by the right hand side of the equation.

The last line shows that $P(k+1)$ is true under the assumption that $P(k)$ is true. This completes the inductive step.

Thus by the Principal of Mathematical Induction, the initial proposition is true for all positive integers n. This completes the proof.

Answer 2

Hint $\ $ Intuitively the induction step arises by applying the Congruence Product Rule

$$ \begin{align}{\rm mod}\,\ 7\!:\qquad \color{#0a0}{12}\ \equiv&\,\ \color{#c00} 5\\ 12^n\equiv&\,\ 5^{n-1}(-2)\qquad\ {\rm i.e.}\ \ P(n)\\\overset{\rm multiply}\Rightarrow\ \ \color{#0a0}{12}\cdot 12^n\equiv&\,\ \color{#c00}5\cdot 5^{n-1}(-2)\quad {\rm i.e.}\ \ P(n\!+\!1)\end{align}\ \qquad $$

Remark $\ $ If we eliminate the language of congruences by substituting inline the linked proof of the Congruence Product Rule then we obtain precisely the common proof in most other answers. I wrote the congruence proof in this form (vs. the more common way) to highlight this relationship.

Thus the inductive step need not be pulled out of a hat like magic. Rather, it has very intuitive arithmetical content as congruence multiplication. See also this answer where I elaborate on this viewpoint (it shows in gory detail precisely how the inductive proof is equivalent to eliminating congruence language by substituting inline the proof of the Congruence Product Rule).

If we exploit to the hilt the congruence viewpoint then applying CPR we can substitute $\, -2\equiv 5\,$ reducing the proof to $\,12\equiv 5\,\Rightarrow\, 12^n\equiv 5^n,\,$ a special case of the Congruence power Rule. This is a prototypical example of the great simplification afforded by the use of congruences in induction problems of this arithmetical type. Often they allow one to reduce the induction to a trivial induction,, e.g. $\,(-1)^{2n}\equiv 1\,$ or $\,1^n\equiv 1,\,$ e.g. the above as $\,12^n/5^n\equiv (12/5)^n\equiv 1^n \equiv 1.$

Generally, unlike above, there is no need to massage the congruence proof into some form to highlight its relationship with another proof. In that case we can simply apply the congruence rules (laws) in whatever manner proves convenient (this is what was done in the linked common congruence proof, where no explicit mention was made when invoking both the congruence laws and the trivial induction conveniently encapsulated in the power law $\,12\equiv 5\,\Rightarrow\,12^n\equiv 5^n).$

Answer 3

For $n=k$, we have $$12^k+2\cdot 5^{k-1} = 7M$$ Multiplying by $12$ throughout, we obtain $$12^{k+1}+24\cdot 5^{k-1} = 84M \implies 12^{k+1} + (2\cdot 5 + 14)\cdot 5^{k-1} = 84 M$$ which in-turn implies $$12^{k+1} + 2\cdot 5^k +14\cdot5^{k-1} = 84M \implies 12^{k+1}+2\cdot 5^k = 14\left(6M-5^{k-1}\right)$$ Hence, the statement is true for $n=k+1$.

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We have $$12^n = (7+5)^n = \sum_{k=0}^{n} \dbinom{n}k 7^k \cdot 5^{n-k} = 5^n + 7 \cdot \sum_{k=1}^{n} \dbinom{n}k 7^{k-1} \cdot 5^{n-k}$$ This gives us \begin{align} 12^n + 2 \cdot 5^{n-1} & = 2\cdot 5^{n-1} + 5^n + 7 \cdot \sum_{k=1}^{n} \dbinom{n}k 7^{k-1} \cdot 5^{n-k}\\ & = \left(2+5\right)\cdot 5^{n-1} + 7 \cdot \sum_{k=1}^{n} \dbinom{n}k 7^{k-1} \cdot 5^{n-k}\\ & = 7 \cdot \left(5^{n-1} + \sum_{k=1}^{n} \dbinom{n}k 7^{k-1} \cdot 5^{n-k}\right) \end{align}

Answer 4

Base step (you didn't mention it): for $n=1$ we have $12^1+2(5^{1-1})=14=7\cdot 2$.

Continuation to your proof:

$12^k(12)+2(5^k)=\underbrace{5(12^k+2(5^{k-1}))}_{\text{inductive hypothesis}}+\underbrace{12^k(7)}_{\text{obvious}}$

Answer 5

You should write $12^k+2\cdot 5^{k-1}=7r$ (using $k$ with two different meanings is wrong).

Then you have $12^k=7r-2\cdot 5^{k-1}$; now \begin{align} 12^{k+1}+2\cdot 5^k &=12\cdot 12^k+2\cdot 5^k\\ &=12\cdot(7r-2\cdot 5^{k-1})+2\cdot 5^k\\ &=7\cdot12r-24\cdot5^{k-1}+10\cdot5^{k-1} \end{align} and you should be able to conclude.

Remember to check the basis of the induction, that is, the case $n=1$.

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Using congruences it is easier, of course: just observe that $12\equiv 5\pmod{7}$, so $$ 12^n+2\cdot 5^{n-1}\equiv 5^n+2\cdot 5^{n-1}\equiv 5^{n-1}(5+2)\equiv 5^{n-1}\cdot7\equiv0\pmod{7} $$

Answer 6

For $n=k$ you have (note the use of variable $m$) $$12^k+2(5^{k-1})=7m\text{, where }m\in\Bbb{N}$$ For $n=k+1$ you have $$\begin{align}12^{k+1}+2(5^k)&=(7+5)12^k+2\cdot5(5^{k-1})\\&=7\cdot12^k+5(12^k+2(5^{k-1}))\\&=7(12^k)+35m\\&=7(12^k+5m)\\&=7S\text{ ,where }S\in\Bbb{N}\end{align}$$

Answer 7

A proof by induction has the following steps:

(1) Show that the statement is true for $n = 1$

(2) Assume that the statement is true for $n = k$, then show that it is true for $n = k + 1$.

(3) Conclude that it is true for all $n \in \mathbb{N}$.

Try carrying out (1) first.

Then for (2), the key is to start with the statement for $n = k + 1$ and try make it look like the statement for $n = k$ plus something else, and then assume that it is divisible by 7 for n = k, and show that this "something else" is also divisible by 7.

In other words, for (2), start with

$12^{k+1}+2(5^{k-1+1}) = 12 \cdot 12^{k}+2(5 \cdot 5^{k-1}) = ...$

Answer 8

Here is a proof with an emphasis on clarity if you so desire:

For $n\geq 1$, let $S(n)$ denote the statement $$ S(n) : 7\mid(12^n+2(5^{n-1}))\Longleftrightarrow 12^n+2(5^{n-1})=7m, m\in\mathbb{Z}. $$ Base case ($n=1$): $S(1)$ says that $7\mid(12^1+2(5^0))$, and this is true.

Inductive step: Fix some $k\geq 1$ and assume that $S(k)$ is true where $$ S(k) : 7\mid(12^k+2(5^{k-1}))\Longleftrightarrow 12^k+2(5^{k-1})=7\ell, \ell\in\mathbb{Z}. $$ To be proved is that $S(k+1)$ follows where $$ S(k+1) : 7\mid(12^{k+1}+2(5^{k}))\Longleftrightarrow 12^{k+1}+2(5^{k})=7\eta, \eta\in\mathbb{Z}. $$ Beginning with the left-hand side of $S(k+1)$, \begin{align} 12^{k+1} + 2(5^k) &= 12\cdot 12^k +2(5^k)\tag{by definition}\\[0.5em] &= 12(7\ell-2(5^{k-1}))+2(5^k)\tag{by $S(k)$, the ind. hyp.}\\[0.5em] &= 12\cdot 7\ell-24(5^{k-1})+2(5^k)\tag{expand}\\[0.5em] &= 12\cdot 7\ell -24(5^{k-1})+10(5^{k-1})\tag{manipulate}\\[0.5em] &= 12\cdot 7\ell-14(5^{k-1})\tag{group like terms}\\[0.5em] &= 7(12\ell-2(5^{k-1}))\tag{factor out a $7$}\\[0.5em] &= 7\eta,\tag{$\eta=12\ell-2(5^{k-1}); \eta\in\mathbb{Z}$} \end{align} we end up at the right-hand side of $S(k+1)$, completing the inductive step.

Thus, by mathematical induction, the statement $S(n)$ is true for all $n\geq 1$. $\blacksquare$