Evaluate $3^x$ from $0$ to $ n$. $\sum_{i=0}^{\log_{3}(n)} 3^i $

Question

I really don't even know where to start, is there some way to interchange the upper limit and the term to sum over?

Answer 1

First of all, for your notation to make sense, the upper limit of the sum must be a non-negative integer, so $n=3^k$, where $k$ is a non-negative integer.

Once that's assumed, your sum is simply a geometric series. Apply the formula for it.

Sum $S = \frac{a(r^n-1)}{r-1} =\frac{1\cdot (3^{(\log_3 n +1)}-1)}{3-1}= \frac 12 (3n-1)$