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Let $a_1 = 3, a_n = a_1\cdot a_2 \cdots a_{n-1}+2$ for $n \ge 2$. Prove that $a_n = 2^n + 1.$

So far what I've been able to work out through induction is:

Base Case: $n=2: a_2 = a_1 + 2 = 5 = 2^2 + 1 = 5$. Good

Induction Case: Assume: $a_n = 2^n + 1.$ Show: $a_{n+1} = 2^{n+1} + 1$

LHS = $a_1\cdots a_n + 2 = a_1\cdots a_{n-1} \cdot a_n + 2 = a_1\cdots a_{n-1}(2^n + 1) + 2 = $

$ = (a_n - 2)(2^n + 1) + 2 = (2^n + 1 - 2)(2^n + 1) + 2 = (2^n - 1)(2^n + 1) + 2 $

$ = 2^{2n} + 1 $

But the RHS = $ 2^{n+1} + 1 $ , which is not equal to the LHS.

Did I do a substitution incorrectly or is something about my general method off? Thanks for the help

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    For a one-line proof, show that $a_{n+1}-2=a_n(a_n-2)$ and deduce that $a_{n+1}-1=$ $____$ (hence the formula $a_n=2^n+1$ is wrong for $n\geqslant3$). – 2017-01-22
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    Check out $a_3 = 3.5+2 = 2^4+1 = 17$ which is not $2^3+1$. Moreover, $a_4 = 3.5.17+2 = 257 = 2^{8}+1$. – 2017-01-22
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    Okay, thank you, this is good because it means the question is wrong. However, its not great that this textbook has an incorrect problem. – 2017-01-22

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It's seems that the problem is wrong. It seems that the sequence has a general form of $a_n = 2^{2^{n-1}} + 1$

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    Would you mind explaining how you found that as its general form. For I would think the general form would be what I found when I simplified the LHS: 2^(2n)+1 – 2017-01-22
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    @Dsmith Well, you started with a wrong assumption, so you get a wrong result for $a_{n+1}$. – 2017-01-22
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    Oh, duh--of course, I was still going along with the problem (which is clearly incorrect). But to be clear, if the problem were correct, my process would also be correct, yes? – 2017-01-22
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    @Dsmith Yeah, your thinking is alright. Unfortunately, the problem was wrong, which led you to thinking you did something wrong. – 2017-01-22