How to guess that $ \sum_{i=1}^{n}3^i = \frac{3}{2}(3^n - 1)$ As in title how do you guess that $ \sum_{i=1}^{n}3^i = \frac{3}{2}(3^n - 1)$?
I have homework about solving recurrence relations and using iterate method I can find that http://www.wolframalpha.com/input/?i=RSolve%5B%7Ba%5Bn%5D+%3D+a%5Bn-1%5D+%2B+3%5En%2C+a%5B1%5D+%3D+3%7D%2Ca%5Bn%5D%2Cn%5D series shown in this example is equal to $ \sum_{i=1}^{n}3^i$, but I am curious how you change it into  $ \frac{3}{2}(3^n - 1)$. I know how to prove it's true, but I don't know how to guess it on my own. Anyone could explain?
 A: Observe that this is a Geometric Series of $n$ terms, the first Term being $=3$ and common ratio $=3$
we can safely use the formula of summation of $n$ terms
A: Hint: $(3-1)(3^{1}+...+3^{n})=(3^{2}+...+3^{n+1})-(3^{1}+...+3^{n})=...$
A: Well... When I look deep inside of me and try to really see how I would guess it, it comes to this anecdote:

An infinite number of mathematicians walk into a bar. 
  The first asks the bartender for a beer. 
  The second asks for half a beer. The third one says he wants a fourth of a beer. 
  The bartender interrupts, puts two beers on the bar and says:
   “You guys need to learn your limits.”

I think, although I cannot be sure, that this is ingrained somewhere deep in the childhood memories of most mathematicians (maybe most people on earth). You learn your fractions, you notice that $\frac{1}{2} + \frac{1}{4} = \frac{3}{4}$, and $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8}$, then you realize that $\sum_{i=1}^{n} \frac{1}{2^i} = 1 - \frac{1}{2^n}$.
Then, when you see a sum like $1 + 3 + \ldots + 3^n$, you already know what to do.
A: Hint: $$3(3^1 + \ldots 3^k) = (3^1 + \ldots 3^k) + 3(3^{k}-1).$$
