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14 Jun 2026, 05:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
49% (02:32) correct
51%
(02:26)
wrong
based on 148
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For each positive integer \(j\), let \(C_j = 3j^2 - 1\). For instance, \(C_1 = 3(1)^2 - 1 = 2\) and \(C_2 = 3(2)^2 - 1 = 11\). For which of the following values of \(n\) does the sum \(C_1 + C_2 + C_3 + ... + C_n\) yield an even number?
I. 24
II. 50
III. 61
(A) I only
(B) II only
(C) I and III only
(D) II and III only
(E) I, II, and III
I. 24
II. 50
III. 61
(A) I only
(B) II only
(C) I and III only
(D) II and III only
(E) I, II, and III
General Discussion
14 Jun 2026, 22:39
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Bunuel
When
- j = even → \(C_j\) = 3 * even - odd = even - odd = odd
- j = odd → \(C_j\) = 3 * odd - odd = odd - odd = even
j = 1 → \(C_j\) = even
j = 2 → \(C_j\) = odd
j = 3 → \(C_j\) = even
j = 4 → \(C_j\) = odd
j = 5 → \(C_j\) = even
j = 6 → \(C_j\) = odd
j = 7 → \(C_j\) = even
j = 8 → \(C_j\) = odd
j = 9 → \(C_j\) = even
The sum of the series \(C_1 + C_2 + C_3 + ... + C_n\)
n = 1 → \(C_j\) = even
h = 2 → \(C_j\) = odd
n = 3 → \(C_j\) = odd
n = 4 → \(C_j\) = even
n = 5 → \(C_j\) = even
n = 6 → \(C_j\) = odd
n = 7 → \(C_j\) = odd
n = 8 → \(C_j\) = even
n = 9 → \(C_j\) = even
Therefore we observe a pattern that the series repeat after every fourth term, after the second term.
Hence, if we represent n = 4x + r
If r = 0 → The sum of the series is odd
Ex: 4 ; 4 * 1 + 0
If r = 1 → The sum of the series is even
Ex: 9 ; 4 * 2 + 1
If r = 2 → The sum of the series is even
Ex: 2 ; 4 * 0 + 2
If r = 3 → The sum of the series is even
Ex: 3 ; 4 * 0 + 3
Now, let's test out for the numbers given -
I. 24
24 = 4 * 6 + 0
Hence, the sum, \(C_1 + C_2 + C_3 + ... + C_{24}\) , will be even
II. 50
50 = 4 * 12 + 2
Hence, the sum, \(C_1 + C_2 + C_3 + ... + C_{50}\) , will be odd
III. 61
61 = 4 * 15 + 1
Hence, the sum, \(C_1 + C_2 + C_3 + ... + C_{61}\) , will be even
Answer: I & III only
Option C
