how do you write the closed form of a sum of the geometric progression of 3^n? Our teacher told us that $2^0+2^1.... 2^n$ is equal to $2^{n+1}-1$ but I am not sure how to apply that to a similar scenario where $3^0+3^1.... 3^n$. I tried $3^{n+1}-1$ but that is not correct. What is the correct way to figure it?
Edit: I know that the summation is $$\sum_{k=0}^{n} 3^k$$ I am looking for a closed form solution.
On
Though the other answers are correct, I figured that if you only know about the sum $1 + 2 + 2^2 + \cdots + 2^{n - 1}$, it is probably the best to write more details and avoid symbols whenever possible. Here is a proof that you can take a look at. Let's denote your sum as $S = 1 + 3 + 3^2 + \cdots + 3^{n - 1}$. Then you want to find $S$ (duh). Now since it is a geometric series, if you multiply $S$ by $3$, you will get
$$ 3S = 3 \times 1 + 3 \times 3 + 3 \times 3^2 + \cdots + 3 \times 3^{n - 1} = 3 + 3^2 + 3^3 + \cdots + 3^n $$
It might not be obvious how this is useful. However, let's compare $3S$ and the original series $S$ as follows:
$$ \begin{array}{lll} & &3S &= & & &3 &+ &3^2 &+ &3^3 &+ &\cdots &+ &3^{n - 1} &+ &3^n \\ &- &S &= &1 &+ &3 &+ &3^2 &+ &3^3 &+ &\cdots &+ &3^{n - 1} \\ \hline \\ & &2S &= &-1 &&&&&&&&&&&+ &3^n \end{array} $$
Where the middle terms all cancel out because you are subtracting! Looking at the last row, we simply get
$$ 2S = 3^n - 1 $$
Which of course means that
$$ S = \frac{3^n - 1}{2} $$
As other suggested. I highly encourage you to copy this proof but replacing $3$ with $5$, or if you are comfortable enough, to replace $3$ by $x$.
In general, we have $$1+x+...+x^n=\frac{1-x^{n+1}}{1-x}$$
You can prove it by Induction as follows- $$1+x+...+x^n+x^{n+1}=\frac{1-x^{n+1}}{1-x}+x^{n+1}=\frac{1-x^{n+2}}{1-x}$$
On second thought, since you have asked the correct way to figure it, one should probably exhibit that the terms cancel out and you get $$(1+x+...+x^n)(1-x)=1-x^{n+1}$$