Problem:
1 , 2 , 1 , 2 , 2 , 1 , 2 , 2 , 2 , 1 , 2 , 2 , 2 , 2 , 1 , 2 , 2 , 2 , 2 , 2 , 1 , 2 , … 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 1 2 2 n 2 n 2 n th n th 1234 1 2 3 4
Answer Choices:
A. 1996 1 9 9 6
B. 2419 2 4 1 9
C. 2429 2 4 2 9
D. 2439 2 4 3 9
E. 2449 2 4 4 9
Solution:
The k th 1 k th 1
1 + 2 + 3 + ⋯ + k = k ( k + 1 ) 2 1 + 2 + 3 + ⋯ + k = 2 k ( k + 1 )
and 49 ( 50 ) 2 < 1234 < 50 ( 51 ) 2 2 4 9 ( 5 0 ) < 1 2 3 4 < 2 5 0 ( 5 1 ) 491 4 9 1 1234 1 2 3 4 2 2 1234 ( 2 ) − 49 = 2419 1 2 3 4 ( 2 ) − 4 9 = 2 4 1 9
OR OR
The sum of all the terms through the occurrence of the k th 1 k th 1
1 + ( 2 + 1 ) + ( 2 + 2 + 1 ) + ⋯ + ( 2 + 2 + ⋯ + 2 ⏟ k − 1 + 1 ) = 1 + 3 + 5 + ⋯ + ( 2 k − 1 ) = k 2 . = = 1 + ( 2 + 1 ) + ( 2 + 2 + 1 ) + ⋯ + ( k − 1 2 + 2 + ⋯ + 2 + 1 ) 1 + 3 + 5 + ⋯ + ( 2 k − 1 ) k 2 .
The k th 1 k th 1
1 + 2 + 3 + ⋯ + k = k ( k + 1 ) 2 . 1 + 2 + 3 + ⋯ + k = 2 k ( k + 1 ) .
It follows that the last 1 1 1234 1 2 3 4 1225 1 2 2 5 k = 49 k = 4 9 1225 1 2 2 5 4 9 2 = 2401 4 9 2 = 2 4 0 1 2 2 18 1 8 2401 + 18 = 2419 2 4 0 1 + 1 8 = 2 4 1 9