As others have mentioned, there are many different possibilities for how the sequence is to be reckoned.
If the sequence is like this: $1, 1 + 3*1, 1+3*1 + 3*2, 1+3*1+ 3*2 + 3*3, ...$
then the $n$th term will be $T_n = 1 + \frac 32n(n-1)$ based on the AP sum, properly adjusted.
If the sequence is like this: $1, 1 + 3*2^0, 1+3*2^0 + 3*2^1, 1+3*2^0 + 3*2^1 + 3*2^2, ...$
then the $n$th term will be $T_n = 1 + 3(2^{n-1} - 1)$, based on the GP sum, properly adjusted.
If (and this possibility hasn't yet been mentioned - but it corresponds to what you guessed in your question text, the sequence is like this: $1, 1+(1+2), 1+(1+2)+(1+2+3), 1+(1+2)+(1+2+3)+(1+2+3+4)+...$,
then the $n$th term will be $T_n = \frac 16n(n+1)(n+2)$, a derivation based on the sum of the first $n$ triangular numbers.
The thing to take home from all this is that a sequence must be more rigorously defined before we can work on it.