Applying Induction Hypothesis

Question

"Prove that for any even number $n\in\mathbb{N}, 2^{3n-1} + 5 \cdot 3^n$ is divisible by $11$"

After proving the base case, I wrote my Induction Hypothesis as such:

Assume $2^{3k-1} + 5 \cdot 3^k$ is divisible by 11 for some even $k\in\mathbb{N}$. Then, we must show $11 | 2^{3(k+2)-1} + 5 \cdot 3^{k+2}$.

These are the steps I took from here,

$2^{3k+5} + 5 \cdot 3^k \cdot 3^2$
$\Rightarrow 2^{3k-1} \cdot 2^6 + 5 \cdot 3^k \cdot 3^2$.

I'm a little stuck here algebraically, I'm not too sure how to apply the Induction Hypothesis in this position.

Answer 1

To match the induction hypothesis, the next step would be

$$2^{3k-1}\cdot 2^6 + 5\cdot 3^k \cdot 3^2 = 2^6\left(2^{3k-1} + 5\cdot 3^k\right) +\left(3^2 - 2^6\right)5\cdot 3^k$$

Answer 2

You can transform the question to :

For all natural numbers $n≥1$, prove that :$$11\mid (2^{6n-1} + 5 \cdot 9^n) $$

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For $n=1$, the statement is obviously correct. To construct the inductive step, we assume that for $n=k$, the statement is correct :

$$2^{6k-1} + 5 \cdot 9^k=11m,\,m\in\mathbb Z$$

Then for $k+1$, you have :

$$ \begin{align}2^{6k+5}+5\cdot 9^{k+1}&=2^{6k+5}+9\left(11m-2^{6k-1}\right)\\ &=99m+2^{6k-1}\left(2^6-9\right)\\ &=99m+55\cdot 2^{6k-1}\thinspace .\end{align} $$

By induction, you are done .