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