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The problem goes like this: Construct a sequence by setting $a_1 = 5$. Then compute $a_3$ and $a_4$ then prove that the $n^{th}$ term is given by the formula $a_n = 3^n+2$.

The sequence goes like this: $a_1, a_2, a_3, a_4, ....,$ [Formula here to calculate $a_n$th term].

a_1 starts at 5 and to calculate $a_2$ we do the following: 3(5) - 4 so for the formula I wrote it as 3($a_n$ - 1) - 4. However, this entire sequence is equal to $3^n+2$ and when using $n = 1$ to show both sides have a base case (proof by induction) it clearly does not work.

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It works ! If $a_1=5$ and $a_n=3a_{n-1}-4$ for $n \ge 2$, then a straightforward inductive proof gives $a_n=3^n+2$.

Try the proof !

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  • $\begingroup$ Oh shoot does it? If n has to be greater or equal to 2 then how could we ever get 5. $\endgroup$ – Zdravstvuyte94 Oct 29 '18 at 13:00

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