hb wrote:
In the sequence 1, 2, 2, …, an, …, an = an-1 • an-2. The value of a13 is how many times the value of a11?
(A) 2
(B) 23
(C) 232
(D) 264
(E) 289
Comments: The n, n-1, n-2, 13, 11 mentioned in the question are subscripts above. I could not figure out how to show them as subscript while writing the formula here.
Disclaimer: I have used the Search Box Before Posting. I used the first sentence of the question or a string of words exactly as they show up in the question below for my search. I did not receive an exact match for my question.
Source: Veritas Prep; Book 04
Chapter: Homework
Topic: Algebra
Question: 93
Question: Page 226
Solution: PDF Page 17 of 18
Edition: Third
My Question: Please provide an explanation on how to arrive at the answer.
The question should read:
In the sequence 1, 2, 2, …, \(a_n\), …, \(a_n = a_{n-1}* a_{n-2}\). The value of \(a_{13}\) is how many times the value of \(a_{11}\)? (A) 2
(B) 2^3
(C) 2^32
(D) 2^64
(E) 2^89
For such kind of questions it's almost always a good idea to write down first terms:
\(a_1=1=2^0\)
\(a_2=2=2^1\)
\(a_3=a_2*a_1=1*2=2^1\)
\(a_4=a_3*a_2=2*2=2^2\)
\(a_5=a_4*a_3=4*2=2^3\)
\(a_6=a_5*a_4=8*4=2^5\)
\(a_7=a_6*a_5=32*8=2^8\)
If you notice exponents form Fibonacci sequence: {0, 1, 1, 2, 3, 5, 8, ...} (Fibonacci sequence is a sequence where each subsequent number is the sum of the previous two)
So, it will continue as follows: {0, 1, 1, 2, 3, 5, 8, 5+8=
13, 8+13=
21, 13+21=
34, 21+34=
55, 34+55=
89, 55+89=
144, ...}
From above we have that \(a_{11}=2^{55}\) and \(a_{13}=2^{144}\).
\(\frac{a_{13}}{a_{11}}=\frac{2^{144}}{2^{55}}=2^{89}\)
Answer: E.