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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

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    $\begingroup$ You've defined a sequence $a_n$; what do you want to prove about it? $\endgroup$ – Christopher Carl Heckman Mar 13 '16 at 23:50
  • $\begingroup$ @CarlHeckman the first step to the question was to determine a formula for the sequence: 3, 6, 12, 24, 48, 96. I did that, but then it asks to prove the formula through induction. I'm not sure how to structure that though $\endgroup$ – Bob Mar 13 '16 at 23:53
  • $\begingroup$ What do they mean by 'prove the formula'? $\endgroup$ – Justin Benfield Mar 13 '16 at 23:56
  • $\begingroup$ If that's what the exact wording was, then your task is impossible. There are infinitely many sequences that start off with $3,6,12,24,48,96$. Did the question have any other details? (Also, you're missing some parentheses in your formula.) $\endgroup$ – Christopher Carl Heckman Mar 13 '16 at 23:57
  • $\begingroup$ ^ Agreed, and that is reason I've never liked these kinds of questions: There are many answers consistent with the given information. For instance, I could give you a polynomial that gives those same 6 numbers as the first 6 terms, but it would not agree with the exponential you've given beyond that. $\endgroup$ – Justin Benfield Mar 13 '16 at 23:59
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Quote from a comment:

the first step to the question was to determine a formula for the sequence: 3, 6, 12, 24, 48, 96. I did that, but then it asks to prove the formula through induction.

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.

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