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26 May 2017, 04:48
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There is a sequence \(S_{n}\) such that \(S_{1} = \{1\}, S_{2} = \{3,5\},\) and \(S_{3} = \{7,9,11\}, \ldots\) . The number of elements in set \(S_{j+1}\) is 1 greater than the number of elements in set \(S_{j, }\) and the smallest element in set \(S_{j+1}\) is 2 greater than the largest element in set \(S_{j}\). What is the value of the largest element of set \(S_{50}\)?
A. 2,509
B. 2,529
C. 2,549
D. 2,551
E. 2,569
A. 2,509
B. 2,529
C. 2,549
D. 2,551
E. 2,569
26 May 2017, 08:39
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Official Solution:
There is a sequence \(S_{n}\) such that \(S_{1} = \{1\}, S_{2} = \{3,5\},\) and \(S_{3} = \{7,9,11\}, \ldots\) . The number of elements in set \(S_{j+1}\) is 1 greater than the number of elements in set \(S_{j, }\) and the smallest element in set \(S_{j+1}\) is 2 greater than the largest element in set \(S_{j}\). What is the value of the largest element of set \(S_{50}\)?
A. 2,509
B. 2,529
C. 2,549
D. 2,551
E. 2,569
If you assume the first term of each \(S_{n}\) as \(a_{n}\), you get \(a_{1} = 1, a_{2} = 3, a_{3} = 7\),
\(a_{4} = 13, \ldots\) ., \(a_{n.}\) In other words, from \(a_{2} - a_{1} = 3 - 1 = 2*1, a_{3} - a_{2} = 7 - 3 = 4 = 2*2, a_{4} - a_{3} = 13 - 7 = 6 = 2*3, \ldots \ldots\) .
you get \(a_{n} - a_{n-1} = 2(n-1)\), then it becomes
\(a_{n} - a_{n-1} = 2(n-1)\)
\(a_{n-1} - a_{n-2} = 2(n-3)\)
..
\(a_{3} - a_{2} = 2*2\)
a_{2} - a_{1} = 2*1\(If you add both sides, respectively, you will get\)(a_{n}-a_{n-1})+(a_{n-1}-a_{n-2})+\ldots .+(a_{3}-a_{2})+(a_{2}-a_{1}) = 2(n-1)+2(n-2)+\ldots+2*2+2*1\(.Then, everything on the left side of the equation is taken out as below.\\
\\
\\
\\
What remains on the right side after the addition is\)a_{n} - a_{1} = 2(n-1+n-2+\ldots +2+1) = \frac{2n(n-1)}{2} = n(n-1)\(, which becomes\)a_{n} = a_{1} + n(n-1) = 1+n(n-1) = n^{2}-n+1\(. Therefore, the last term of\)S_{50}\(is 2 less than the first term of\)S_{51}\(, so the first term of\)S_{51} = 51^{2} - 51+1 = 2,551\(and the last term of\)S_{50} =\(the first term of\)S_{51}-2 = 2,551 - 2 = 2,549$. Therefore, the answer is C
Answer: C
There is a sequence \(S_{n}\) such that \(S_{1} = \{1\}, S_{2} = \{3,5\},\) and \(S_{3} = \{7,9,11\}, \ldots\) . The number of elements in set \(S_{j+1}\) is 1 greater than the number of elements in set \(S_{j, }\) and the smallest element in set \(S_{j+1}\) is 2 greater than the largest element in set \(S_{j}\). What is the value of the largest element of set \(S_{50}\)?
A. 2,509
B. 2,529
C. 2,549
D. 2,551
E. 2,569
If you assume the first term of each \(S_{n}\) as \(a_{n}\), you get \(a_{1} = 1, a_{2} = 3, a_{3} = 7\),
\(a_{4} = 13, \ldots\) ., \(a_{n.}\) In other words, from \(a_{2} - a_{1} = 3 - 1 = 2*1, a_{3} - a_{2} = 7 - 3 = 4 = 2*2, a_{4} - a_{3} = 13 - 7 = 6 = 2*3, \ldots \ldots\) .
you get \(a_{n} - a_{n-1} = 2(n-1)\), then it becomes
\(a_{n} - a_{n-1} = 2(n-1)\)
\(a_{n-1} - a_{n-2} = 2(n-3)\)
..
\(a_{3} - a_{2} = 2*2\)
a_{2} - a_{1} = 2*1\(If you add both sides, respectively, you will get\)(a_{n}-a_{n-1})+(a_{n-1}-a_{n-2})+\ldots .+(a_{3}-a_{2})+(a_{2}-a_{1}) = 2(n-1)+2(n-2)+\ldots+2*2+2*1\(.Then, everything on the left side of the equation is taken out as below.\\
\\
\\\\
What remains on the right side after the addition is\)a_{n} - a_{1} = 2(n-1+n-2+\ldots +2+1) = \frac{2n(n-1)}{2} = n(n-1)\(, which becomes\)a_{n} = a_{1} + n(n-1) = 1+n(n-1) = n^{2}-n+1\(. Therefore, the last term of\)S_{50}\(is 2 less than the first term of\)S_{51}\(, so the first term of\)S_{51} = 51^{2} - 51+1 = 2,551\(and the last term of\)S_{50} =\(the first term of\)S_{51}-2 = 2,551 - 2 = 2,549$. Therefore, the answer is C
Answer: C
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