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Let $a_1 = 4$. Let, for each natural number $n$, $a_{n+1} = 3a_n-8$. Devise an explicit formula for $a_n$, and finally use induction to prove it.

Upon cursory inspection, I noticed that $a_2 = 3(4) - 8 = 4$. This clearly creates an infinite loop; $a_3 = 4\cdots$.

Am I missing something? Am I supposed to prove the formula $a_n=4$ by induction? Is this a typo?

It should be noted that this problem was sourced from a textbook, so an error is not out of the question.

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Gothenburgresident
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    Yes, you are meant to prove that $a_n=4$ inductively. As you remark, this is not difficult. – lulu Aug 29 '21 at 20:00

1 Answers1

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Yes the statement can be formally proved by induction, notably:

  • base case: $a_1=4$
  • induction step: assume as hypothesis $a_n=4$ then

$$a_{n+1}=3a_n-8\stackrel{Ind. Hyp.}=3\cdot 4-8=12-8=4$$

and the proof is complete.

Note that the infinite loop you refer to, that is $a_1=4 \implies a_2=4 \implies a_3=4 \implies \cdots$, is indeed the heart of induction idea (which is usually stated as an axiom of the natural numbers).

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