The Series Series

• I'll divide this solution in two parts to help you folks understand
• I : We can distribute this sequence in $\boxed{1}, \boxed{2, 1}, \boxed{2, 2, 1}$ ... .
• If we pay attention enough, we can easily see we have 1 term in the first box, 2 terms in the second one, 3 in the third, and this keeps going, forming an arithmetical progression. So, we can say we have $\frac{((1 + x)x}{2} = 1234$ .
• Solving for that, we'll find x is slightly bigger than 49 . This means we have 49 boxes and an incomplete one, filled with 2s .
• Repeating $\frac{((1 + x)x}{2}$ with x = 49 , we'll find 1225 as answer, with means we have $1234 - 1225 = 9$ 2s from the incomplete box; if we sum them, we'll have $9\times2 = 18$ . Let's keep this information for a while.

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• II : Back to the 49 boxes, we have that the sum of the first box terms equals 1, the sum of the second box terms equals 3, the sum of the third box terms equals 5, and again it keeps going forming another arithmetical progression.
• We already have the sum of the first box terms equals 1, and we can state that sum of the last box terms is $1 + 48\times2 = 97$ .
• This way, we can afford that the sum of all terms in boxes equals $\frac{49\times(1 + 97)}{2} = 2401$ .
• Finally, adding 2401 to our previously-found 18 , we can finally conclude that $S_{1234}$ = 2419 .

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• Hey, I know this solution got I little bigger and maybe complex, so, if you have any doubt or question or comment, just ask bellow and i'll do my best to help.

the sequence is: 1,2,1,2,2,1,2,2,2,1,2,2,2,2,1,2,2,2,2,2....
Notice that there are two 2s for the second 1
three 2s for the third 1
......and so on
=> 1234= x{1s} +1+2+3+4+....x {2s}
=> 1234= x{1s}+ x{x+1}/2 {2s}
solving we get x = 48 approx
thus when x=48, no. of terms= 48+48{49)/2
=1224
=>there are 10 more terms
=>no. of 1=48+1=49
& no. of 2s=48(49)/2 +9
=> sum=49*1 + 1185 * 2 =2419