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I'm pretty much stuck and I'm obviously missing some hint to come up with a proof for the following problem:

Given the series

$$\ a_0 =1$$ $$\ a_n =4(a_{n-1}+1)$$

I want to show that $a_n\equiv 1\pmod 7 $ for all $n$

I calculated these:

$\ a_0 =1, a_1=8, a_2=36, a_3 =148$

The base is clear, we take $n = 1$, then we get $$\ a_1 = 4(a_{n-1}+1) = 4(a_{1-1}+1)=4(a_{0}+1)=4(1+1)=8$$

Afterwards i continued with:

$n \to n+1$

$$a_{n+1} = 4(a_{n}+1)$$

We can replace $a_n$ with $4(a_{n-1}+1)$

$$a_{n+1} = 4(4(a_{n-1}+1)+1)$$ or $$a_{n+1} = 16(a_{n-1})+20$$

How should I proceed from here on? Is there another way to prove this problem? I see that it is recursive - I'm just kind of stuck.

Thanks in advance for any help :)

Bill Dubuque
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1 Answers1

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$a_n \equiv 1 \pmod{7} \implies (a_n + 1) \equiv 2 \pmod{7} \implies $

$a_{n+1} = 4(a_n + 1) \equiv 1 \pmod{7}.$

The idea is that $8 \equiv 1\pmod{7}.$

user2661923
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