Question

The sequence 1,2,1,2,2,1,2,2,2,1,2,2,2,2,1,2,2,2,2,2,1, etc. consists of 1s separated by blocks of 2's with n 2's in the nth block. Find the sum of the first 1234 terms in the sequence.

Answer 1

2418 i guess

Answer 2

if you have 1234 2's you'd have 2468, but since there's about a 50 1's in there as well, you have to subtract the 1's

Answer 3

Subtract the ones? Wouldn't they also add to the total sum?

Answer 4

if you assume all the 1234 numbers are two, then you have to subtract the ones, because you have counted the 1s as 2s, and since a 2 is bigger then a 1, you'll have to get rid of the extras

Answer 5

Ah

Answer 6

So how many ones are there?

Answer 7

not sure exactly how many, but 50 should be about right probably

Answer 8

you get a whole lot of 2s so there's hardly any room for 1s past the first few

Answer 9

i tried to make a formula but i can't think of one

Answer 10

I got 2419 instead

Answer 11

f(n) = 1 if n = i(i+1)/2, and 2 if n = otherwise

Answer 12

1234 = i(i+1)/2

when solving, the nearest integer less than 1234 is 1225, which makes i = 49.

So there are 49 1's, and 1185 2's

(49)(1) + (1185)(2) = 2419