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0

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


2

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


0

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


0

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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