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The original question is the following

prove that $2·\sum_{i=0}^{n-1} 3^{i} = 3^n-1$ for all n $\geq$ 1

I know that I have to prove by induction and have successfully done the base case, my IH is the following:

Assume $2·\sum_{i=0}^{n-1} 3^{i} = 3^n-1$ holds for all n $\geq$ 1 to prove $2·\sum_{i=0}^{n} 3^{i} = 3^{n+1}-1$

then I did this: $2·\sum_{i=0}^{n} 3^{i} = 2·\sum_{i=0}^{n-1} 3^{i} +3^n$

then by my IH: $2·\sum_{i=0}^{n-1} 3^{i} +3^n = 3^n-1+3^n $

after this I struggle to continue, I know I have to find a way to rewrite it to $3^{n+1}-1$ and I know this is equal to $3·3^n-1$ but I don't see how to get from the one to the other. Anyone who can help?

edit: I mistook a 2 for a 3, can anyone help now?

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    $\begingroup$ Perhaps the question is meant to be to prove that $2·\sum_{i=0}^{n-1} 3^{i} = 3^n-1$ ? $\endgroup$ – gandalf61 Jan 13 at 12:34
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    $\begingroup$ ... and now your mistake is clear. When moving from $2·\sum_{i=0}^{n-1} 3^{i}$ to $2·\sum_{i=0}^{n} 3^{i}$ you need to add $2·3^n$, not just $3^n$. $\endgroup$ – gandalf61 Jan 13 at 15:34
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You can add the first pair of paranthesis in the derivation below, so that it becomes clear that the factor $2$ also multiplies the second term, as follows:

$2 \cdot \sum_{i=0}^n 3^i = 2 \cdot \left(\sum_{i=0}^{n-1} 3^i + 3^n \right) = 2 \cdot \sum_{i=1}^{n-1} 3^i + 2 \cdot 3^n = (3^n -1 ) + 2 \cdot 3^n = 3^{n+1}-1.$

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$2·\sum_{i=0}^{n} 3^{i} = 3^n-1 +2\cdot{3^n}=3^{n+1}-1$.

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Considering that the statement is incorrect (indeed $2\cdot\sum\limits_{i=0}^{n-1}2^i +1 =\sum\limits_{i=0}^{n}2^i = 2^{n+1}-1\neq 3^n$), I doupt anyone here can prove it.

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As you can find as the basis of induction, it is not true even for $n = 2$ (Easy to see $2(1‌+2) = 6 \neq 3^2 - 1 = 8$).

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