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The sequence 1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2,...., consists of 1's separated by blocks of 2's with n 2's in the nth block.
What is the sum of the first 1234 terms of this sequence?
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Lets group the sequence into the sum of segments
S =
(1+2) +
(1+2+2) +
(1+2+2+2) +
(1+2+2+2+2) +
. . . . . . . . . . (1)
The sum inside each k-th segment is (1+2k), k = 1, 2, 3, 4, 5, . . .
The lengths of the segments form the sequence 2, 3, 4, 5, . . .
So the lengths of segments make the arithmetic progression with the first term 2 and the common difference 1.
If we have 48 segments, their accumulated length is
2 + 3 + 4 + 5 + . . . + 49 = = 1224. ( ! <<<---+++ As "everybody" knows, 1 + 2 + 3 + 4 + 5 + . . . + n = )
So, I will find the sum (1) for 48 segments, such that the number of terms in this sum will be 1224.
I think that the problem MUST ask about the sum of 1224 terms, not 1234.
I believe that your 1234 is a TYPO, while 1224 is the TRUE value.
Then the sum of the first 1224 terms
s = 3 + 5 + 7 + 9 + . . . + (2*48+1) = 49^2 - 1 = 2400. ( ! <<<---+++ As "everybody" knows, 1 + 3 + 5 + 7 + 9 + . . . + (2n-1) = )
Answer. The sum of 1224 first terms of the given sequence is 2400.
Regarding these famous sums, see the lessons on arithmetic progressions
- Arithmetic progressions
- The proofs of the formulas for arithmetic progressions
- Problems on arithmetic progressions
in this site.
Also, you have this free of charge online textbook in ALGEBRA-II in this site
- ALGEBRA-II - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic
"Arithmetic progressions".