# Thread: proof by induction 5^n-1 is divisible by 4

1. ## proof by induction 5^n-1 is divisible by 4

Use mathematical induction to prove that the assertion is true for $\displaystyle n\ge 1$. $\displaystyle 5^n-1$ is divisible by $\displaystyle 4$.

basis $\displaystyle n_{o}=1$ then $\displaystyle 5^1-1=4$ is divisible by $\displaystyle 4$.

IH let $\displaystyle 1\le l<k$ and assume $\displaystyle 5^l-1$ is divisible by $\displaystyle 4$.

Induction Step
$\displaystyle 5^{k}-1=(5)5^{k-1}-1$
Since $\displaystyle 5^{k-1}-1$ is divisble by $\displaystyle 4$ (by the induction hypothesis) then $\displaystyle (5)5^{k-1}-1$ is also divisible by $\displaystyle 4$

Is this right?
What is wrong with the latex

2. ## Re: proof by induction 5^n-1 is divisible by 4

The induction step should have n = k+1

3. ## Re: proof by induction 5^n-1 is divisible by 4

IH let $\displaystyle 1\le l<k$ and assume $\displaystyle 5^l-1$ is divisible by $\displaystyle 4$.
If this is a proof by induction you need to go a step at a time, not prove it for all numbers at once.

I'd say this. "Assume $\displaystyle 4 |(5^k-1)$. We want to show $\displaystyle 4|( 5^k^+^1-1)$."

Since $\displaystyle 5^{k-1}-1$ is divisble by $\displaystyle 4$ (by the induction hypothesis) then $\displaystyle (5)5^{k-1}-1$ is also divisible by $\displaystyle 4$
Not quite, no. You have the formula wrong, and you need to show your work; how do you know 4 divides $\displaystyle 5^k^+^1-1$ for certain?

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# 5^n-1 is divisible by 4

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