I'm currently learning mathematical induction, but I'm not sure how to start a proof for this formula(the problem is different than what I've been practicing with). I'm not looking for the whole solution I'm just stumped on how to structure the proof for the following formula: $a_n = 3\cdot 2^n-1$, for all integers $n \geq 1$.
Any help would be greatly appreciated
Quote from a comment:
The stated formula, $a_n = 3\cdot 2^n-1$, is incorrect: If $n=1$ then $3\cdot2^n-1 = 3\cdot2-1 = 6 - 1 = 5$, and $5$ is not the first number in the sequence.
However, $a_n = 3\cdot 2^{n-1}$ is correct, for $n=1$ through $6$.
You can prove that simply by computing all six numbers. Induction usually involves proving an infinite sequence of statement. I don't know if I've ever seen it used for a finite sequence. However, in order to prove this by induction, you would need some rule that the sequence is known to follow other than the one you're trying to prove is right, and the only one available is the one you're trying to prove is right. So this cannot be done by induction, but it can be done by checking the six cases.