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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Hit Kudus if this has helped you get closer to your goal, and also to assist others save time. Tq