Consider the lengths of consecutive 1-2 blocks.
block 1 - 1, 2 - length 2
block 2 - 1, 2, 2 - length 3
block 3 - 1, 2, 2, 2 - length 4
block 4 - 1, 2, 2, 2, 2 - length 5
and so on.
Recall the formula for the sum of consecutive positive integers,

Now,

which means that the 1234th term in the sequence occurs somewhere about 1/5 of the way through the 49th 1-2 block.
In the first 48 blocks, the sequence contains 48 copies of 1 and 1 + 2 + 3 + ... + 47 copies of 2, hence they make up a total of

numbers, and their sum is

This leaves us with the contribution of the first 10 terms in the 49th block, which consist of one 1 and nine 2s with a sum of
.
So, the sum of the first 1234 terms in the sequence is 2419.