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?