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If a sequence is defined by an = a(n-1)*a(n-2)
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07 Apr 2022, 05:43
07 Apr 2022, 05:43
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Question Stats:
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If a sequence is defined by \(a_n = a_{n-1}*a_{n-2} + 1\) for \(n \geq 3\) and if \(a_1 = 1\) and \(a_2=1\), what is the value of 6th term?
(A) 1
(B) 7
(C) 22
(D) 155
(E) 721
(A) 1
(B) 7
(C) 22
(D) 155
(E) 721
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Posts: 6218
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Re: If a sequence is defined by an = a(n-1)*a(n-2)
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07 Apr 2022, 05:50
07 Apr 2022, 05:50
Carcass wrote:
If a sequence is defined by \(a_n = a_{n-1}*a_{n-2} + 1\) for \(n \geq 3\) and if \(a_1 = 1\) and \(a_2=1\), what is the value of 6th term?
(A) 1
(B) 7
(C) 22
(D) 155
(E) 721
(A) 1
(B) 7
(C) 22
(D) 155
(E) 721
Given: \(a_n = a_{n-1} \times a_{n-2} + 1\)
Interpretation: Each term is equal to 1 greater than the product of the two terms before it.
For example: \(a_6 = a_{5} \times a_{4} + 1\)
As you can see, in order to find \(a_6\), we must first find the values of \(a_5\), \(a_4\) and \(a_3\)
Let's begin calculating the terms after \(a_2\)
\(a_1 = 1\)
\(a_2=1\)
\(a_3 = a_{2} \times a_{1} + 1 = 1 \times 1 + 1 = 2\)
\(a_4 = a_{3} \times a_{2} + 1 = 2 \times 1 + 1 = 3\)
\(a_5 = a_{4} \times a_{3} + 1 = 3 \times 2 + 1 = 7\)
\(a_6 = a_{5} \times a_{4} + 1 = 7 \times 3 + 1 = 22\)
Answer: C
gmatclubot
Re: If a sequence is defined by an = a(n-1)*a(n-2) [#permalink]
07 Apr 2022, 05:50
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