Question

The sequence$1,2,1,2,2,1,2,2,2,1,2,2,2,2,1,2,2,2,2,2,1,2,\dots$consists of $1

Answer 1

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,

$\displaystyle \sum_{i=1}^j i = 1 + 2 + 3 + \cdots + j = \frac{j(j+1)}2 \implies \sum_{i=2}^j = \frac{j(j+1) - 2}2$

Now,

$1234 = \dfrac{j(j+1)-2}2 \implies 2470 = j(j+1) \implies j\approx49.2016$

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

$\displaystyle \sum_{i=1}^48 1 + \sum_{i=1}^{48} i = 48+\frac{48(48+1)}2 = 1224$

numbers, and their sum is

$\displaystyle \sum_{i=1}^{48} 1 + \sum_{i=1}^{48} 2i = 48 + 48(48+1) = 48\times50 = 2400$

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 $1+9\times2=19$.

So, the sum of the first 1234 terms in the sequence is 2419.

Answer 2

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,

$\displaystyle \sum_{i=1}^j i = 1 + 2 + 3 + \cdots + j = \frac{j(j+1)}2 \implies \sum_{i=2}^j = \frac{j(j+1) - 2}2$

Now,

$1234 = \dfrac{j(j+1)-2}2 \implies 2470 = j(j+1) \implies j\approx49.2016$

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

$\displaystyle \sum_{i=1}^48 1 + \sum_{i=1}^{48} i = 48+\frac{48(48+1)}2 = 1224$

numbers, and their sum is

$\displaystyle \sum_{i=1}^{48} 1 + \sum_{i=1}^{48} 2i = 48 + 48(48+1) = 48\times50 = 2400$

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 $1+9\times2=19$.

So, the sum of the first 1234 terms in the sequence is 2419.

Answer 3

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,

$\displaystyle \sum_{i=1}^j i = 1 + 2 + 3 + \cdots + j = \frac{j(j+1)}2 \implies \sum_{i=2}^j = \frac{j(j+1) - 2}2$

Now,

$1234 = \dfrac{j(j+1)-2}2 \implies 2470 = j(j+1) \implies j\approx49.2016$

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

$\displaystyle \sum_{i=1}^48 1 + \sum_{i=1}^{48} i = 48+\frac{48(48+1)}2 = 1224$

numbers, and their sum is

$\displaystyle \sum_{i=1}^{48} 1 + \sum_{i=1}^{48} 2i = 48 + 48(48+1) = 48\times50 = 2400$

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 $1+9\times2=19$.

So, the sum of the first 1234 terms in the sequence is 2419.

Answer 4

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,

$\displaystyle \sum_{i=1}^j i = 1 + 2 + 3 + \cdots + j = \frac{j(j+1)}2 \implies \sum_{i=2}^j = \frac{j(j+1) - 2}2$

Now,

$1234 = \dfrac{j(j+1)-2}2 \implies 2470 = j(j+1) \implies j\approx49.2016$

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

$\displaystyle \sum_{i=1}^48 1 + \sum_{i=1}^{48} i = 48+\frac{48(48+1)}2 = 1224$

numbers, and their sum is

$\displaystyle \sum_{i=1}^{48} 1 + \sum_{i=1}^{48} 2i = 48 + 48(48+1) = 48\times50 = 2400$

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 $1+9\times2=19$.

So, the sum of the first 1234 terms in the sequence is 2419.