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.
(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.