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Difficulty:
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Question Stats:
37% (02:14) correct
63%
(02:20)
wrong
based on 60
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Let \(a_1\), \(a_2\), .., be a sequence with the following properties.
I. \(a_1 = 1\), and
II. \(a_{2n} = n*a_n\) for any positive integer n.
What is the value of \(a_{2^{100}}\)
A. 1
B. \(2^{99}\)
C. \(2^{100}\)
D. \(2^{4950}\)
E. \(2^{9999}\)
sequences.JPG [ 18.41 KiB | Viewed 4311 times ]
I. \(a_1 = 1\), and
II. \(a_{2n} = n*a_n\) for any positive integer n.
What is the value of \(a_{2^{100}}\)
A. 1
B. \(2^{99}\)
C. \(2^{100}\)
D. \(2^{4950}\)
E. \(2^{9999}\)
Attachment:
sequences.JPG [ 18.41 KiB | Viewed 4311 times ]
02 Apr 2022, 17:26
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twobagels
As the question is written in what I quote above, the answer is just 1, because \(a_2 = a_{2(1)}\), so according to property II, since n = 1 we have \(a_2 = 1*a_1 = 1\). So (a_2)^100 = 1^100 = 1.
But in the spoiler tag, there's a screenshot of the question, and it says something different. The exponent in the question is part of the subscript, so we want the value of the 2^100th term in the sequence (not the value of the second term raised to the power 100). Again using property II, since 2^100 = (2)(2^99), we'll have n = 2^99, so we learn that
\(\\
a_{2^{100}} = a_{(2)(2^{99})} = 2^{99} \times a_{2^{99}}\\
\)
and similarly, since 2^99 = (2)(2^98), continuing the above, we have
\(\\
a_{2^{100}} = a_{(2)(2^{99})} = 2^{99} \times a_{2^{99}} = 2^{99} \times a_{(2)(2^{98})} = 2^{99} \times 2^{98} \times a_{2^{98}}\\
\)
and if we keep doing this, we'll just end up with
\(\\
(2^{99})(2^{98})(2^{97}) \ldots (2^2)(2)(1) = 2^{99 + 98 + 97 + \ldots + 2 + 1}\\
\)
and since we're adding 99 terms with an average of 50, this equals \(2^{99 \times 50} = 2^{5000 - 50} = 2^{4950}\)
What is the source?
02 Apr 2024, 12:11
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