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If a_n = (a_{n-1} - 1)(a_{n-1} + 1) and a_1 = 1, what is the value of

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Expert Post
Math Expert
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V
Joined: 02 Sep 2009
Posts: 42338

Kudos [?]: 133122 [0], given: 12415

If a_n = (a_{n-1} - 1)(a_{n-1} + 1) and a_1 = 1, what is the value of [#permalink]

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New post 12 Sep 2017, 05:25
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E

Difficulty:

  55% (hard)

Question Stats:

62% (00:59) correct 38% (01:06) wrong based on 50 sessions

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Kudos [?]: 133122 [0], given: 12415

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Joined: 02 Nov 2015
Posts: 176

Kudos [?]: 26 [0], given: 121

GMAT 1: 640 Q49 V29
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Re: If a_n = (a_{n-1} - 1)(a_{n-1} + 1) and a_1 = 1, what is the value of [#permalink]

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New post 12 Sep 2017, 06:56
It's an E..
Every even number will be 0 and every odd number -1.

So answer is -1.

Sent from my Lenovo TAB S8-50LC using GMAT Club Forum mobile app

Kudos [?]: 26 [0], given: 121

Manager
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Kudos [?]: 54 [0], given: 341

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Re: If a_n = (a_{n-1} - 1)(a_{n-1} + 1) and a_1 = 1, what is the value of [#permalink]

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New post 12 Sep 2017, 07:58
Bunuel wrote:
If \(a_n = (a_{n-1} - 1)(a_{n-1} + 1)\) and \(a_1 = 1\), what is the value of \(a_{25}\)?


A. 25
B. 5
C. 0
D. 1
E. –1


simplify the equation
a_n = (a_{n-1} -1)(a_{n-1} + 1)
this is in the form (X-Y)(X+Y)= X^2 - Y^2
a_n = a_{n-1}^2 - 1^2
a_2 = {a_1}^2 -1 = 1-1 =0
a_3 = {a_2}^2 -1 = 0-1 =-1

likewise if you continue, you will observe the pattern that gives 0 for even terms and -1 for odd terms.
a_25 is an odd term and hence equals -1.
"E"
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Kudos [?]: 54 [0], given: 341

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Kudos [?]: 917 [0], given: 5

Re: If a_n = (a_{n-1} - 1)(a_{n-1} + 1) and a_1 = 1, what is the value of [#permalink]

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New post 15 Sep 2017, 10:16
Bunuel wrote:
If \(a_n = (a_{n-1} - 1)(a_{n-1} + 1)\) and \(a_1 = 1\), what is the value of \(a_{25}\)?


A. 25
B. 5
C. 0
D. 1
E. –1


We see that a(2) = 0 x 2 = 0, a(3) = -1 x 1 = -1, a(4) = -2 x 0 = 0, a(5) = -1 x 1 = -1, ...

We see that for n > 1, a(n) = 0 if n is even and a(n) = -1 if n is odd.

Thus, a(25) = -1.

Answer: E
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Jeffery Miller
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Kudos [?]: 917 [0], given: 5

Re: If a_n = (a_{n-1} - 1)(a_{n-1} + 1) and a_1 = 1, what is the value of   [#permalink] 15 Sep 2017, 10:16
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