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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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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

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Posts: 163
GMAT 1: 640 Q49 V29 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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It's an E..
Every even number will be 0 and every odd number -1.

So answer is -1.

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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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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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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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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.

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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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_________________ Re: If a_n = (a_{n-1} - 1)(a_{n-1} + 1) and a_1 = 1, what is the value of   [#permalink] 30 Mar 2019, 20:23
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