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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1051
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B_{n} = (−1)^n × n + 3 for all integers n ≥ 1. What is the 9th term of
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03 Feb 2021, 09:25
03 Feb 2021, 09:25
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\(B_{n} = (−1)^n × n + 3\) for all integers n ≥ 1. What is the 9th term of the sequence?
(A) -9
(B) -6
(C) -1
(D) 6
(E) 9
(A) -9
(B) -6
(C) -1
(D) 6
(E) 9
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Re: B_{n} = (−1)^n × n + 3 for all integers n ≥ 1. What is the 9th term of
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03 Feb 2021, 09:55
03 Feb 2021, 09:55
1
9th term of the sequence will be nothing but \(B_{9}\) (as n ≥ 1)
So, to find \(B_{9}\) we need to substitute n=9 in \(B_{n} = (−1)^n × n + 3\)
=> \(B_{9} = (−1)^9 × 9 + 3\)
=> \(B_{9} = (−1)× 9 + 3\)
=> \(B_{n} = -9 + 3\) = -6
So, answer will be B
Hope it helps!
So, to find \(B_{9}\) we need to substitute n=9 in \(B_{n} = (−1)^n × n + 3\)
=> \(B_{9} = (−1)^9 × 9 + 3\)
=> \(B_{9} = (−1)× 9 + 3\)
=> \(B_{n} = -9 + 3\) = -6
So, answer will be B
Hope it helps!
Re: B_{n} = (1)^n × n + 3 for all integers n 1. What is the 9th term of
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16 Apr 2024, 06:16
16 Apr 2024, 06:16
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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