The partial sum of a geometric sequence is
[tex]\displaystyle \sum_{i=0}^N a^i = \dfrac{a^{N+1}-1}{a-1}[/tex]
In your case a=3, so if we sum N terms of the sequence we have
[tex]\displaystyle \sum_{i=0}^N 3^i = \dfrac{3^{N+1}-1}{2}[/tex]
We want this to me more than 1 million, so we have
[tex]\dfrac{3^{N+1}-1}{2}>1000000 \iff 3^{N+1}-1>2000000 \iff 3^{N+1} > 1999999[/tex]
Considering the log (base 3) of both sides, we have
[tex]N+1>\log_3(1999999)\iff N>\log_3(1999999)-1 approx 12.2[/tex]
So, starting from N=13, the sum of the first N terms will be more than 1 million
